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TEXTILE 


Applied Arithmetic 

r AND 

Calculations 




W 


c. 


BY 

W. McSWAIN 






Copyright 19 19 

By Victor-Monaghan Company 

Published October, 19 19 



ffEC:l7i3l9 



©CI.A5610 20 



Foreword 



This work was undertaken after a thorough inves- 
tigation of the best means of providing education 
for the man who had either been deprived of this 
advantage, or who had been so employed as to have 
become dulled by not having to apply an early 
education to his daily task. 

Realizing that the best way to carry out this cam- 
paign was to consult those who would afterward 
benefit from this research,, the author arranged con- 
ferences with superintendents, overseers, second 
hands, section men and other employes. At these 
meetings it was found that the form of education 
most needed was that bearing on the necessary 
calculations appertaining to the textile industry. 
It was, however, found that, in most cases, the men 
realized that such calculations could not be suc- 
cessfully studied until a better knowledge of mathe- 
matics was obtained. The result was the idea of 
incorporating the two in one. Hence this book. 

This book being written for and at the instigation 
of cotton mill men, it is well adapted to their use. 
While repetition may occur at several places, it 
is intended to be so, and is believed that it enhances 
rather than depreciates its value. Only such prin- 
ciples of mathematics are used as are needed to 
fully understand the methods used in textile cal- 
culations. 

The author wishes to express his thanks to and 
appreciation of the valuable aid given by officials 



and employes of the Victor-Monaghan Company in 
compiling this book. Many of the problems incor- 
porated were submitted by some of the employes 
of this Company, and many valuable suggestions 
were given by others. 

C. W. McSWAIN. 



DEDICATION 

V 

To those men and boys, employed in the 
Cotton Mills of the South, who desire to 
fit themselves to become successful in this, 
our most important industry, this book is 
dedicated. 

— The Author. 



Numbers 



A number is an expression showing how many 
times an object or thing is taken. If we see an 
object and then another of the same general physical 
or mechanical construction we say that there are 
two of these objects,, and if we see still another 
object of the same make-up we say three and so on 
for as many times as we conceive in our mind that 
this thing occurs at a given place or in a given time. 

The symbols used in writing numbers are com- 
monly called figures. These figures which we use 
to denote numbers were brought into use about two 
thousand years ago. They were brought from India 
by the Arabs and are still called Arabic or Hindu- 
Arabic numerals. There are nine of these figures 
besides zero which is sometimes called naught or 
cipher. These figures are 1, 2, 3, 4, 5, 6, 7, 8, 9, 0. 

In common practical uses there are three different 
kinds of numbers all using the same figures in ex- 
pressing their values. These are : Whole Numbers, 
Fractional Numbers or Fractions, and Decimal Num- 
bers or Decimals. 

WHOLE NUMBERS 

"When we write figures to show the number of 
times a certain thing is taken as a whole these 
figures are called whole numbers. Thus, if we say 
"two hundred bales of cotton", we write the figures, 
200, which means that we are considering this many 
separate individual bales of cotton, all of which 
have the same general make-up in regard to their- 



physical composition as well as to their mechanical 
construction. 

In writing large numbers it is found convenient 
to separate the whole series of figures into groups 
of three ; thus, 250,615,112. By this means it is 
easier to read the whole by considering the first 
group of three to the right as in hundreds place. 
Thus„ to read this part of the number we say "one 
hundred and twelve." The second group to the 
left of this is in thousands place and is read "six 
hundred and fifteen thousand. ' ' The third group 
to the left of this one is in millions place and is read 
"two hundred and fifty million." Thus, the whole 
of this number is read: "two hundred and fifty 
million, six hundred and fifteen thousand, one hun- 
dred and twelve." 

Exercise 

"Write in figures : 

1. Fifty bales of cotton. 

2. One hundred and twelve rolls. 

3. Six hundred and one spindles. 

4. One thousand and one pounds. 

5. Two thousand, one hundred and nine ounces. 

6. Two hundred and fifteen thousand, eight hun- 
dred and twenty-five grains. 

7. Forty-five flyers. 

8. One hundred and one yards of roving. 

9. Two hundred and fifty-three pounds of card 
strippings. 

10. One thousand, six hundred and twelve spin- 
ning rings. 

11. Seven hundred and eighty-five reeds. 

12. Ten thousand, eight hundred and six heddles. 



13. Six hundred and five thousand, eight hun- 
dred and two minutes. 

FRACTIONS 

When any one whole thing is divided into several 
parts one of these parts is called a fraction. Thus, 
if a bale of cotton weighing 400 pounds is divided 
into two parts each of these parts will be 200 pounds 
and is called one half of the whole. If this 400 
pound bale of cotton is divided into 4 equal parts 
each of the four parts will weigh 100 pounds and is 
called one-fourth of the whole. Thus we see that 
each of the parts when the whole is divided into 
four parts is exactly one-half of those parts when 
the division was made in 200 pound lots. There- 
fore the number one-fourth is one-half as large as 
the number one-half. 

The figures used to denote a fraction are placed 
one above a line and one below, thus, £ (one-half), 
| (three-fourths), -J (one-eighth), f (three-eighths), 
y iG (one-sixteenth), %c (three-sixteenths). The 
figure below the line, called the Denominator, shows 
into how many parts the whole number has been 
divided. Thus -J means that the whole was divided 
into two parts, \ that it was divided into four parts, 
■§ into eight parts, and Y 16 into sixteen parts. 

The figure above the line, called the Numerator, 
shows how many of these parts are taken. Thus 
if we divide a number into eight equal parts each 
of these parts is called one-eighth (|), and if we 
desire to show that three of these parts are to be 
taken we write the figure three above the line, 
thus: f (three-eighths). 

All fractions are divided into three general 



classes, known as proper fractions, improper frac- 
tions, and mixed fractions. 

PROPER FRACTIONS 

When the number above the line (numerator) is 
smaller than the number below the line (denom- 
inator) the whole is said to be a proper fraction. 

Thus the following are all proper fractions : •§■ (one- 
eighth), | (three-eighths), $• (four-fifths), % (three- 
sixths), f (two-thirds), | (three-fifths), % (six- 
eighths), % (five-eighths), % 6 (four-sixteenths) r 
15 / 1Q (fifteen-sixteenths), 17 / 2 o (seventeen-twen- 
tieths). 

IMPROPER FRACTIONS 

When the number above the line (numerator) 
is greater than that below the line (denominator) 
the whole is said to be an improper fraction. Thus, 
the following are improper fractions : % (five- 
fourths), % (six-halves), % (seven-thirds), % (nine- 
fifths), % (six-fourths), xx /-i (eleven-sevenths). 

MIXED NUMBERS 

When a whole number precedes or is set before 
a fraction the whole is said to be a mixed number. 
Thus, the following are mixed numbers: 5£ (five 
and one-half), 4| (four and three-fourths), 10f (ten 
and three-eighths), 25^ (twenty-five and one-fourth), 
llG 1 ^ (one hundred and sixteen and eleven-six- 
teenths), HOf (one hundred and ten and five- 
eighths). 



10 



Exercise 

Write in figures: 

1. One-half a revolution of a roll. 

2. Three-eighths of a bale of cotton. 

3. Eleven-sixteenths is the diameter of this pipe. 

4. Two and seven-sixteenths is the diameter of 
this shaft. 

5. Eleven-fifths is equal to two and one-fifth. 

6. The diameter of this roll is one and seven- 
eighths inches. 

7. Eight and one-half revolutions of the cord 
cylinder. 

DECIMALS 

Decimal notation is a method of expressing the 
values of numbers which are less than one in a de- 
creasing ratio of 10. If, for instance, one inch be 
divided into ten equal parts, one of these parts is 
called "one-tenth" of one inch. This is expressed 
decimally by placing a point to the left of the 
number, thus : .1. If each of these ten divisions be 
divided in turn into ten more equal parts the total 
number of divisions in one inch would be 100 and 
each one of these parts would be called "one- 
hundredth" and is expressed decimally thus .01. In 
like manner if we again divide each of these latter 
divisions into ten more, the total number would be 
one thousand and each of these divisions would be 
"one-thousandth" and is written thus : .001. If then 
we desire to show how many of each of these parts 
are to be taken we proceed as follows : If we take 
2 parts when one inch is divided into 10 parts it is 
read: "two tenths" and is written .2. If we take 

11 



3 parts it is read: ''three tenths" and is written .3. 
If we take four parts it is read "four tenths" and is 
written .4. If we take five parts or one-half of the 
total number of parts it is read "five-tenths" and 
is written .5. When we proceed thus for all ten 
parts we arrive at the whole or "ten-tenths" and 
it is written 1.0. Again when the total number of 
divisions is one hundred,, one of these divisions is 
called "one hundredth" and is written .01. If two 
divisions are taken it is read "two one-hundredths " 
and is written .02, and so on for the total number 
which brings us back to "one tenth". 

In like manner do we proceed in writing and 
reading the parts when the whole is divided into a 
thousand parts. For instance, if four of such 
divisions be taken it is read "four one-thousandths" 
and is written .004. 

Exercise 

Write in figures : 

1. Three-tenths of a mile. 

2. Twenty-five hundredths is equal to £. 

3. Seventy-five hundredths is equal to f. 

4. Two and five-tenths equals 2\. 

5. Three and f ourteen-hundredths multiplied by 
the diameter equals distance round a circle. 

6. Seven and eighty-five hundredths is the cir- 
cumference of a circle which is 2| inches in diameter. 

7. Two-tenths of a day. 

8. Eight and seven-tenths revolutions of a cal- 
endar roll. 

9. Six and eighty-seven hundredths yards of, 
warp. 

12 



Iii common practice decimals are read by stating 
first where the point is placed and then giving the 
figures following in order. Thus .5416 is read: 
' ' point, five, four,, one, six ' \ 

Write in figures : 

1. Point, one, two, three, six. 

2. Two, point, four, six, eight. 

3. Ten, point, six,, eight, two. 

4. One hundred and twelve, point, six, five. 



13 



Addition of Whole Numbers 



In adding small numbers, such as those which are 
smaller than 10, it is sometimes customary to place 
them in a line with the sign of addition, (+, plus), 
between them. Thus, 2+4+3+5=14, is read, "two 
plus four, plus three, plus five equals fourteen". 
When large numbers are added they are usually 
placed one below the other so that the last figure on 
the right for each number is directly under the 
same corresponding figure in the number above. 
Thus to add 204, 16, 1012 and 12002 we proceed as 
follows : 

204 

16 

1012 

12002 



13234 



The process above is to add first the row of figures 
at the right, namely, 2, 2, 6 and 4. This gives 14 
and the figure 4 is placed below the line directly 
below the set of figures thus added and the figure 1 
is added into the next line of figures to the left. 
This is continued until all rows have been thus added 
and the resulting number below the line is the sum 
of all the numbers thus added. 

CARD ROOM PROBLEMS 

1. Ten bales of cotton are opened having the 
following weights : 514 lbs., 476 lbs., 504 lbs., 524 

14 



lbs., 487 lbs., 497 lbs., 510 lbs., 507 lbs., 494 lbs., and 
502 lbs. What is the total weight? 

2. Six laps are weighed as follows : 42 lbs., 40 
lbs., 43 lbs., 42 lbs., 41 lbs., 40 lbs. What is the 
total weight? 

3. If a man received 18 dollars per week, how 
much would he have received at the end of four 
weeks ? 

"4. A mill has 2 slubbers with 92 spindles each, 
2 slubbers with 96 spindles each, and 4 slubbers 
with 88 spindles each. What is the total number of 
slubber spindles in mill? 

SPINNING ROOM PROBLEMS 

1. Six doffs are made on spinning frames having 
following weights : 52 lbs., 51 lbs., 50 lbs., 53 lbs., 
51 lbs., 52 lbs. What is the total weight of yarn 
made on the six frames? 

2. A Spinning Room produces 8572 lbs. of 30 's 
yarn, 10,514 lbs. of 24 's, and 6451 lbs. of 28 's. What 
is the total weight of yarn made? 

3. If a Spinning Room makes 5416 lbs. of 32 's, 
11,562 lbs. of 30 's, and 4154 lbs. of 24 's in one week, 
how many pounds of 32 's should it make in 4 weeks ? 
(Add 5416 four times.) How many pounds of 30 's 
should it make in three weeks? (Add 11,562 three 
times.) How many pounds of 24 's should it make 
in five weeks? (Add 4154 five times.) 

4. If a girl runs 10 sides of 150 spindles each 
how many spindles does she tend? If each spindle 
will produce (theoretically) 2 pounds in one week, 
each side should make 300 pounds. How many 
pounds should the 10 sides make? (Add 300 10 
times. ) 

15 



WEAVE ROOM PROBLEMS 

1. A loom has twelve harness on each of which 
are the following numbers of heddles : 416,, 420, 424, 
418, 82, 82, 106,, 106, 116, 116, 46, 46. What are the 
total number of heddles on all harnesses? 

2. A Weave Room produces the following num- 
bers of cuts in a week : Monday, 142 ; Tuesday, 
264; Wednesday, 250; Thursday, 246; Friday, 232; 
Saturday, 164. What is the total number of cuts? 

3. Eight warper beams are placed behind the 
slasher weighing respectively 448 lbs., 446 lbs., 450 
lbs., 447 lbs., 448 lbs., 446 lbs., 451 lbs. What is 
the total weight of yarn unsized? 

4. If eight warper beams, each having 416 ends, 
are behind the slasher, what is the total number of 
ends on each loom beam? (Add 416 eight times.) 



1G 



Subtraction of Whole Numbers 



Subtraction is the process of finding the differ- 
ence between two unequal numbers. The sign of sub- 
traction is made thus : — and is called minus. When 
this sign is placed between two numbers, it means 
that the latter one is to be taken, or subtracted, 
from the former. 

Subtraction may be considered as an inverted 
form of addition, because when a smaller number 
is taken from a larger the resulting number when 
again added to the smaller will equal the larger. 
Thus 4—2=2. This is the same as saying that the 
difference between 4 and 2 is the same as whatever 
number is required to be added to 2 to make 4. 

In subtracting large numbers they are placed with 
the smaller one below the larger the same as in 
addition. Then by starting at the right side the 
first figure in the bottom number is subtracted from 
the one above. If the top figure is smaller than the 
bottom the number 10 is added to the top, making 
a temporary number of either 10, 11, 12, 13, 14, 15, 
16, 17,. or 18 according as to whether the top number 
is either 0, 1, 2, 3, 4, 5, 6, 7, or 8. When this is done 
for a row of figures then to the next row to the 
left the bottom figure must be increased by "one" 
before proceeding with the subtraction. 

Example : Subtract 2168 from 3415. 

3415 
-2168 

1247 

' 17 



Ill the above example we take the first two num- 
bers to the right in bottom and top number, and 
as 5 is smaller than 8 we mentally add 10 making 
16. Then we say "8 from 15= (equals) 7,." and 
place this number, 7, below the line. Then before 
subtracting the next two figures add 1 to the bottom 
figure, 6. This makes 7, and we again proceed as 
before,, saying "7 from 11 is 4", and place this 
figure below the line. In like manner Ave proceed 
subtracting all the figures in the set. In this case 
the third set of figures contains 4 above and 1 below, 
hence it is not necessary to call the top number 14. 
As, however, we added 10 to the top number in the 
set before this one we must add 1 to the bottom 
figure. This gives 2 and we say "2 from 4 equals 
2" and place the 2 below the line. In the fourth 
set of figures it is not necessary to add 1 to the 
bottom number as Ave did not add 10 to the top in 
the last set. Therefore Ave say "2 from 3 equals 1" 
and place this beloAv the line. The result is 1247 
which is the difference betAveen 3415 and 2168. This 
can be proved by adding 1247 and 2168 Avhich will 
give the original number 3415 as follows: 

1247 

2168 



3415 

CARD ROOM PROBLEMS 

1. If ten bales of cotton each Aveighing respec- 
tively 487, 515, 520, 492, 498, 522, 508, 494, 498, 522, 
508, 494, 496 and 512 pounds has 247 pounds of 
bagging and ties Avhat is the net (Aveight after 
deducting bagging and ties) Aveight of cotton? 

18 



2. Locally ginned cotton has bagging and ties 
to the amount of 22 pounds. Compressed cotton 24 
pounds. If 8 bales of locally ginned cotton weigh 
respectively 492, 498, 502, 504, 498, 496, 492 and 498 
pounds, and 6 bales of compressed cotton -weighing 
respectively 512, 506, 508,. 498, 502 and 504 pounds 
are opened, what is the net weight of cotton opened? 
(Add 22 together 8 times for total weight of bagging 
and ties on the eight bales of cotton locally ginned. 
Subtract this weight from total weight of all 8 
bales. Then add 24 together six times to find 
weight of bagging and ties on the 6 bales of com- 
pressed cotton. Subtract this weight from the total 
weight of the 6 bales of compressed cotton. Add 
together the difference in each case and the result 
in the total number of pounds of cotton opened.) 

3. If cotton is opened to the amount of 20,000 
lbs. and from this weight 18,642 lbs. of laps are 
made, how much waste Avould you expect in the 
Picker Room? Suppose that 1162 lbs. of waste is 
all that can be accounted for, what has become of 
the remainder and how much is it? 

4. If 4 laps weighing 40 lbs. each are put through 
the card and 152 lbs. of sliver is delivered, what is 
the amount of waste? 

SPINNING ROOM PROBLEMS 

1. If a mill has 31,652 spinning spindles and 
12,840 are on warp yarn, what is the number of 
filling spindles? 

2. 62,847 pounds is the weekly production of 
certain spinning room. If 38,117 pounds of this is 
warp, how much is filling? 

19 



3. An order is received for 100,000 lbs. of 28 's 
yarn. 12,450 lbs. are produced the first week, 11,872 
lbs. are produced the second week, and 15,117 lbs. 
the third week. How much remains to be made? 

4. A mill has 1492 looms, 216 of these are stand- 
spindles stopped for one week, how many spindles 
run? 

WEAVE ROOM PROBLEMS 

1. An order for 50,000 pieces is received and 
8752 are made the first week, 7985 the second week 
and 9415 pieces the third week. How many pieces 
remain to be made? 

2. If 8 warp beams each weighing respectively 
432, 434,. 435, 432, 432, 433 and 434 lbs. are run on 
a set, on slasher and loom beams are made totaling 
3683 lbs. together with 9 lbs. of waste, how many 
pounds of size are put on? 

3. A cut of cloth weighs 14 pounds. If it has 
8 pounds of warp in it, what is the weight of the 
filling? 

4. A mill has 38,252 spindles. If there are 2560 
ing for a week, how many looms are runnings 



20 



Multiplication of Whole Numbers 

The process of taking a certain one number for 
a particular number of times is called multiplica- 
tion. If we consider the number 2 taken three 
times the result is 6. Therefore we may consider 
multiplication as the process of adding together a 
certain number for a particular number of times. 
Thus, if we add 3 to itself three times we have 
3-f-3+3= (equals) 9. As we have taken this num- 
ber together three times to make 9, we might say 
"3 taken three times equals 9," or "3 times 3 equals 
9." In like manner if we desire to multiply one 
number by another it is not necessary for the per- 
son unskilled in figures to worry himself about 
trying to remember all of the multiplication tables. 
Of course as multiplication is simply a short form 
of addition for each individual number it is more 
convenient for one who expects to. become proficient 
in mathematics to learn by practice the result of mul- 
tiplying all of the numbers from 1 to 9 by any of 
these same numbers. It is recommended that all who 
are not familiar with the multiplication tables prac- 
tice this process until it is thoroughly mastered. 

Usually when two small numbers are to be multi- 
plied together they are placed one in front of the 
other with the sign of multiplication, (X), between 
them. Thus if we desire to multiply 6 by 4 we place 
them thus,, 6X4, and say "6 multiplied by 4" or 
"6 times 4." In arriving at the product or result 
of multiplying these two numbers together we may 
proceed in the following manner : If 6 is to be 

21 



multiplied by 4 it means that 6 is to be added to- 
gether four times and we might say ' ' 6 plus 6 plus 6 
plus 6 equals 24, ' ' which is the same thing as saying 
that "6 taken 4 times equals 24" or that 6X4= 
(equals) 24. 

When multiplying a large number by a smaller 
one, or two large numbers together, the usual prac- 
tice is to place one (usually the smaller one) below 
the other as in addition or subtraction. After this 
is done start with the first figure to the right in the 
top number and multiply this by the first figure to 
the right in the bottom number. Remember always 
that "zero" or "naught" or the figure "0" Avhen 
multiplied by any other produces "zero" or 
"naught" or the figure 0. 

Suppose we desire to multiply 1463 by 502. In 
this case we proceed as follows : 

1463 
502 



2926 

0000 (usually omitted.) 
7315 



734426 



We take in this example the first figure, 3, in 
the top number, 1463, and multiply it by the first 
figure, 2, in the bottom number, 502. This gives 6 
and we place it below the line directly below the 
figure, 2. Then proceed to multiply all of the fig- 
ures in the top number by this figure, 2. Thus, 2 
times 6 equals 12. Place the figure 2 below the line 
to the left of 6 and add the figure 1 to the next figure 

22 



produced by multiplying. Thus, 2 times 4 equals 8 
and 1 added gives 9. Then 9 is placed below the 
line to the left of 2 previously obtained. Next the 
figure 1 is multiplied by 2 and as the previous prod- 
uct was not greater than 9 no figure is to be added to 
this one and the result is 2, which is placed to the 
left of 9. This first step then produces the number 
2926 as seen below the line. 

The next step is to multiply again all of the figures 
in the top number by the second figure from the 
right in the bottom number. As, however, this fig- 
ure is "zero" or the result Avould be zero. 

Then proceed to multiply all figures in the top 
number again by the third figure, 5, in the bottom 
number, 502. First, 5 times 3 equals 15. Place 
the 5 directly in line and under the figure, 5, by 
which we are multiplying, and add 1 into the next 
product. Thus, again, 5 times 6 equals 30 and 1 
added gives 31. Place the figure 1 to the left of the 
5 previously obtained and add the 3 into the product. 
Thus 5 times 4 equals 20 and the 3 gives 23. Place 
the 3 to the left of 1 previously obtained and add 2 
into the next product. Thus, 5 times 1 equals 5 and 

2 added gives 7. Place this 7 to the left of the figure 

3 previously obtained. This completes the process 
of multiplying. The next step is to add together 
the numbers produced by each process. In this case 
we have only the numbers 2926 and 7315 to add, but 
the number 7315 is placed with its figure 5 three 
numbers to the left of the first figure, 6, in the num- 
ber 2926. This happens because the third number 
which would have been produced by multiplying by 
the second figure in the bottom number produced 
zero. If this second figure had been any other fig- 

23 



ure than zero we would have had a third number in 
our final adding. Adding then all the vertical rows 
of figures we find the number 734,426, which is the 
product of multiplying together 1463 and 502. 

In multiplying any number by 10 all that is suffi- 
cient is to annex one cipher or zero or to the num- 
ber multiplied. .Thus 416X10=4160; 510x10= 
5100; 1604X10=16040, etc. 

In multiplying any number by 100 all that is 
necessary is to annex two "ciphers" or 0's. Thus, 
116X100=11600; 510X100=51,000; 1004X100= 
100,400, etc. 

In multiplying any number by 1000 all that is 
necessary is to annex three "ciphers" or 0's. Thus, 
26X1000=26,000; 41x1000=41,000; 510X1000= 
510,000, etc. 

EXERCISE 



Multiply 8X9 


17X8 


16X4 


1006X40 


25X4 


410X20 


50X2 


120X60 


126X10 


140X50 


115X15 


132X66 


304X8 


201X80 


510X16 


80X15 



In multiplying any numbers by 20, 30, 40, 50, 60, 
70, 80, or 90 simply multiply the number by either 
2, 3, 4, 5, 6, 7, 8, or 9 and annex one '"naught" to 
the result. Example: Multiply 1240 by 50. 
Thus 

1240 

5 Annexing the ' ' cipher, ' ' 0, we have 62000 

6200 

24 



CARD ROOM PROBLEMS 

1. If a lap weighs 14 oz. per yd., how many 
grains will there be in 1 yard? (1 oz.=437+ grains.) 

2. If a 54 grain sliver is being made with a 98 
draft in card,, how many grains in one yd. of lap? 
(Wt. in grain of sliver X draft = wt. in grain of 
hi p.) 

3. If there are 840 yards in 1 lb. of a No. 1 hank 
roving how many yards would there be in 10 lbs of 
hank roving? 

4. If there are 840 yards in 1 lb. of No. 1 hank 
roving how many yards would there be in 1 lb. of 
No. 6 hank roving? 

5. If a roll is 4 inches in circumference and runs 
116 revolutions per minute (R. P. M.) how many 
inches of roving would be delivered in 1 minute ? In 
1 hour? In 1 day? In 1 week? 

6. If the slubber clock registers 8 hanks per spin- 
dle made in a day, how many hanks would be made 
on a slubber of 92 spindles ? How many yds. of rov- 
ing would this be of 840 yards=l hank? 

SPINNING ROOM PROBLEMS 

1. If there are 840 yards of No. 1 yarn in 1 lb., 
how many yards of No. 1 would there be in 6 lbs. ? 

2. If there are 840 yards of No. 1 yarn in 1 lb., 
how many yards of No. 20 yarn would there be in 
lib.? 

3. If one spindle will produce 6 pounds of No. 8 
yarn in one week, how many pounds ought 11640 
spindles make in the same time? 

4. If the front roll on a spinning frame is 3 
inches in circumference (around) and is running 120 



25 



revolutions per minute, how many inches of yarn 
will be delivered in 10 hours? 

5. The weight of yarn on a full bobbin with 6 
inch traverse is about 2 ounces from a If inch ring. 
How many ounces should a doff weigh from a 32C 
spindle frame? 

6. If a mill has 516 frames with 336 spindles each 
what is the total number of spindles? 

WEAVE ROOM PROBLEMS 

1. If cloth is to weigh 5 yards per pound, how 
many yards would there be in 12 pounds? 

2. If cloth has 64 warp ends in one inch how 
many ends would there be in the warp if cloth is 
36 in. wide? 

3. If a reed has 32 dents on one inch how many 
dents would be on 38 inches? 

4. If a loom makes 160 picks per minute how 
many picks does it make in one hour? In 10 hours? 

5. How many inches of filling would be taken up 
in 60 picks of shuttle if the reed spread of warp is 
38 inches? (One pick would take up 38 inches.) 

6. There are 840 yards of No. 1 yarn in one 
pound. How many yards of No. 1 yarn would there 
be in 6 pounds? How many yards of No. 20 yarn 
would there be in one lb. ? 

7. If a loom makes 42 yards of cloth in one day, 
how many yards would 15 looms make? 

8. If 8 section beams each with 438 ends are up 
behind the slasher what is the' total number of ends 
going on the loom beams? 

9. Sixty-four looms are to be put on a style of 
goods requiring 26 bars in the pattern chain. How 
many bars would there be on all 64 looms? 

26 



10. A pattern requires 8 harness with the fol- 
lowing numbers of heddles on each harness : 1, 432 ; 
2, 438 ; 3, 434 ; 4, 438 ; 5, 82 ; 6, 82 ; 7, 64 ; 8, 64. If 
24 sets of harness are to be put on, how many hed- 
dles would be required in all? 



27 



Division of Whole Numbers 



Division is the process of splitting up a number 
into several parts so that when all of these parts are 
added together the result obtained is the original 
number. Thus, in the number 2 there are two l's 
and we say "1 will go into 2 two times." In like 
manner if we divide 15 into three equal parts each 
one of these parts will be the number, 5. So there 
are three fives in 15 and we say "3 into 15 will go 
5 times." In all division the number we " divide 
into" is always equal to the product of multiplying 
together the number we "divide by" and the num- 
ber obtained by the division. Thus in dividing 20 
by 5 we mean that we desire to divide 20 into 5 
equal parts, so we mentally decide what number 
multiplied by 5 will give 20. This number is 4 and 
we say "5 into 20 will go 4 times," or "20 divided 
by 5 equals 4." 

In dividing small numbers they are usually placed 
one in front of the other with the sign of division 
(-=-) between them. Thus if we desire to divide 
20 by 5 we have 20-^5= (equals) 4. Also 15-^-3 
= (equals) 5; 36-1-6= (equals) 6; 10-^5= (equals) 
2; 20^-2= (equals) 10, etc. 

In some cases a number is indicated to be divided 
by another by placing the first above a line and the 
second below that line. Thus if we have 15 it 

3 

means that 15 is to be divided bv 3. In like man- 



28 



ner we may have.. 20 15 

— = (equals) 4; — = (equals) 3: 
5 5 

16 

— = (equals) 4, etc. 
4 

Iu dividing a large number by a small number, 
(usually one less than 10) it is done by a process 
known as "Short Division." If, for instance, we 
desire to divide 27612 by 6 the process is as follows : 

6127612 



4602 



In this example Ave first divide 6 into the first 
figure on the left. If this figure is smaller than the 
number we are dividing by, we then take the first 
two figures and mentally divide into' these. In this 
case 6 would not go into two, so we take 27 and 
divide 6 into it ; 6 into 27 goes 4 times and 3 over. 
Place the 4 below the line and putting together the 
remainder, 3, and the next figure, 6,, we have 36. 
Then 6 into 36 goes 6 times even, and the 6 is placed 
below the line. Next 6 will not not go into 1, so 
Ave place the figure below the line and run 6 into 
the two figures 1 and 2 making 12. Then 6 goes 
into 12 tAvo even times and the figure 2 is placed 
beloAv the line. 

In dividing one large number by another it is 
usually done by a process knoAvn as "Long Di- 
vision." If, for example, Ave desire to divide 28567 
by 162 the f olloAving process is used : 



29 



162|28567|176 
~162 



1236 
1134 

1027 
972 



55 over 

In this process the number, 162, is placed to the 
left, as seen, and is divided into as many figures on 
the left of large number as is required for it to go 
into. In this case 162 could not be divided into 2, 
nor 28, so we take the three figures, 2, 8, and 5, mak- 
ing 285. Then we try to find out how many times 
162 will go into this number. It must be a number 
which, when 162 is multiplied by it, the result must 
not be greater than the 285. If we were to try 2, 
we would find that 162 when multiplied by 2 would 
give 324. So the first number in the result must 
be 1. Then 162 is multiplied by this first number, 
1, and the result placed under and subtracted from 
the 285. This remainder is 123. Then annex to 
this remainder the next figure in the number being 
divided. In this case it is 6 and we have the num- 
ber 1236, into which we again divide 162. This we 
find will go 7 times and we again multiply 162 by 7 
and place this result, 1134, under the number 1236 
and again subtract. This step gives a remainder of 
102 and we draw down the next figure, 7, from the 
number being divided. This gives 1027, into which 
we again divide 162 and obtain 6. Again 162 mul- 
tiplied by 6 gives 972 and this subtracted from the 
number, 1027, leaves a remainder of 55. Then the 



30 



result of our division is that 162 goes into 28567, 176 
times with 55 over. 



EXERCISE 



Solve, 



10- 


■r-2= 


30- 


r-3= 


40- 


r-10= 


16 




4 




18 




2 




110 




10 





Divide 1672 by 2 
1480 by 4 
1600 by 8 
" 21462 by 213 



Solve 840 yds. 

14 yds. 
" 1140 inches 



yds. 



= yds. 



36 inches 



CARD ROOM PROBLEMS 

1. If there are 840 yds. in 1 pound of No. 1 yarn, 
how many pounds would there be in 11760 yds. of 
the same size yarn? 

2. If a No. 1 hank roving has 840 yds. in 1 pound, 
what would be the hank roving number if there are 
2520 yds. in 1 pound? 

The actual draft of a card is found by dividing 
the weight in grains of one yd. of sliver into weight 
in grain of 1 yd. of lap. 

3. If a card is running a 12 oz. lap and 54 grain 
sliver, what is its draft? 

Reduce the weight in ozs. of the lap to grains. 
There are 437+ grains in 1 oz. Then 12x437=5244 
grains and 



31 



54 1 5244 1 97 +dr aft 
486 

384 
378 



4. If a 54 grain card sliver, when doubled six on 
drawing frames produces a 55 grain sliver, what is 
the draft of drawing ? 

54 X 6=324 
324^55=draft. 

5. What is the weight of sliver produced on a 
drawing frame doubling 6, 50 grain card slivers into 
one with a draft of 5? 

50X6=300 
and 300-f-5= 60 grain sliver. 

6. If there are 36 inches in 1 yd.,, how many yds. 
would there be in 10,188 inches? 

,7. If a Coiler Calendar roll is 7 inches in cir- 
cumference (around), how many yds. of sliver will 
it deliver in 1 hour if it is turning 168 revolutions 
per minute? 

If a roll is seven inches in circumference, when 
it makes one turn 7 inches of sliver will be delivered. 
36 inches 1 yd. 

8. If the Coiler Calendar roll in the above ex- 
ample will deliver 19,600 yds. of a 54 grain sliver 
in 10 hours, what is the production in pounds ? 7000 
grains=l pound. 

9. If the circumference of front roll on slubber 
is 4 inches and it is making 138 revolutions per 

32 



minute, Iioav many hanks of 840 yds. each will be 
delivered in 10 hours' running? If the frame has 
96 spindles, what is the total number of hanks made? 
If the roving is 1 hank, it means that there is 1 hank 
in 1 pound. How many pounds would this be for 
96 spindles? 

SPINNING ROOM PROBLEMS 

1. If there are 840 yds. of a No. 1 yarn in one 
pound, how many pounds would 16,800 yds. weigh 
of the same number? 

2. If a No. 1 yarn has 840 yds. in 1 pound, what 
would be the number of the yarn if there were 25 ; 200 
yds. in 1 pound? 

3. If two ends of an 8 hank roving are run be- 
hind spinning frame, what is the resulting hank 
roving number? Hank roving -f- doublings = hank 
roving fed. 

4. If a 6 hank roving is doubled on spinning 
frame and is being spun into a No. 30 yarn, what is 
the draft in spinning frame? 

6^2=3 and 30^-3=10 draft. 

5. If a frame has a draft of 8 and is making 
24 's yarn, what is the number of hank roving used, 
if it is being used double ? 

24-^8=3, then if double, 2X3=6 hank roving. 

6. If 3 pounds of yarn has 51640 yds., what is 
its number, (840 yds. in 1 pound of number 1) ? 

840X3=2520, then 51640-^2520=No. yarn. 

7. The number of a yarn is found practically by 
reeling 120 yds. and weighing in grain. The weight 
in grain divided into the number 1000, will give the 
number of the yarn. 

33 



If 120 yds. of yarn weigh 50 grain, what is its 
number? 

If 120 yards of yarn weigh 100 grain, what is its 
number? 

If a yarn is No. 20, what should it weigh in grain? 

8. If the weekly production of a spindle on 10 's 
yarn is 2 pounds, how many spindles must be put 
on to deliver a weekly production of 18500 lbs ? 

1 spindle=2 pounds, therefore the number of 

pounds produced is twice the number of spindles, 

therefore the number of spindles would be \ the 

number of pounds, or 18500 

— : =9250 spindles. 

2 

WEAVE ROOM PROBLEMS 

1. If a warp has- 2592 ends in it and cloth made 
from it is 36 inches wide, how many ends per inch 
has the cloth? 

2. If 4 harness are to be used on a warp of 2800 
ends with equal number of ends for all harness, how 
mand heddles should be on each harness? 

3. If a warp 2460 ends is drawn 2 ends per dent 
in reed, how many dents would be taken up in reed? 

4. If a warp requires 1230 dents in reed to be 
drawn and it is to be set 30 inches wide, how many 
dents per inch should there be in the reed? 

5. If five yards of cloth weigh 1 pound, how 
many pounds should 1 cut of 60 yards weigh? 

6. If cloth is to be 36 inches wide, 72 ends per 
inch and woven with fancy pattern carrying 108 
ends to one repeat, how many repeats of pattern 
would there be across cloth? 

7. If cloth is to weigh 4 yds. per pound, what 

34 



would one yard weigh in ounces? (16 ounces==l 
pound.) 

8. If a loom makes 160 picks per minute weaving 
cloth with 40 picks per inch, how many inches of 
cloth would it make for each minute? How many 
inches in 1 hour? (60 minutes.) 

How many inches in 10 hours? How many yards 
w»uld that be in 10 hours? 

9. No. 1 yarn means that there are 840 yds. of 
that size yarn in 1 pound. In No. 2 yarn there are 
2X840 yards in 1 pound and so on for any number 
of yarn taken. 

If we have 25200 yds. of yarn in 1 pound, what 
number is it? 

25200-^840. 

10. If the spread in the reed is 39 inches and the 
picks per inch 48, how many inches of filling would 
there be in one inch of cloth? How many yards 
would this be? How many yards of filling would 
this give in one yard of cloth? 

11. A certain loom makes 40 yards per day. 
How many cuts of 60 yards will it make in 6 days : 

40 yds. = 1 day. In 6 days,, 6X40=240 yds, 
then 240-^60=4 cuts. 

12. A weekly production of 500 cuts of 60 yds. is 
desired from looms which can produce 40 yds. in 10 
hours. How many looms must be put on? 

500X60=30,000 yds. required in 60 hours. 

If 1 loom makes 40 yds. in 10 hours, in 60 hours it 
will make 40X6=240. 30,000-=-240==number of 
looms. 



35 



Fractions 

REDUCTION OF FRACTIONS 

For practical methods all proper fractions are 
reduced to their lowest terms before being used in 
the process. When a whole number is divided into 
two parts, (equal), each of these two parts is called 
one-half, % : and when divided into four parts, 
(equal), each of these parts is called one-fourth, J. 
If we compare these two divisions we will see that 
it will require 2 of the one-fourth divisions to equal 
one-half of the total number, while it will require 
only one of the one-half. divisions to make an equal 
amount. Then we see that the number, %, is the 
same in value or is equal to the number, %■ I n 
like manner % is equal to % or to %. Thus we 
see that multiplying both the top and the bottom 
figures in a fraction by any number, the value of 
the fraction is not changed. Also dividing both the 
top and the bottom figure by a common number does 
not alter its value. Thus if we multiply % by 2, 
we have % ; by 3, %; by 4, %; by 5, % ; by 6, 6 / 12 , 
etc., etc. 

Also if we divide % 6 , both top and bottom, by 4, 
we have % ; by 2, %, and if we divide 1 % 2 by 4 Ave 
have %; by 8, %; by 16, y 2 , etc., etc. 

Any improper fraction may be reduced to a mixed 
number by dividing the bottom figure into the top. 
Thus, iy 5 ==2y 5 , i 2 / 7 =l%, 16 / 5 =3i/ 5 , 22/ 8=2 %,=23/ 4 , 
etc., etc. 

36 



Example, 66 

4 

4 1 66 J 16% 

~~4_' 

26 
24 

% 
*\ mixed number can be reduced to an improper 
fraction by multiplying the whole number by the 
number below the line and adding in the number 
above the line. Use this new number above the 
line and place the old number below the line. Thus, 
to reduce 11% to an improper fraction we have 
6X11=66, 66+5=71. Then the new number will 

y 6 . 



be, ra 



37 



Addition of Fractions 



In adding two or more fractions it is necessary 
to reduce them all to the same term. We cannot 
add Ys and % because the two numbers are express- 
ed in different terms and if we add the two in this 
form, the sum will be neither in "Thirds" or 
"Fourths." In order then to add two fractions it 
is necessary to reduce them to similar fractions. 
This means that all the fractions must be reduced 
so that the numbers under the line for all of them 
are the same. Thus it is simple to add Ys, % and 
% as they are all in the same term or "Eighths." 
This is done by adding together all the figures 
above the line and placing the sum above a new 
line under which must be the same figure which 
occurs below the line of all numbers thus added. 
The sum then of Ys and % and %=%=% (by divid- 
ing top and bottom by 2). 

If,, however, it becomes necessary to add such 
fractions as Ys and Y% and Y15 all must be reduced 
to like terms. Thus, to add Ys> Y% an( l Y15 we mul- 
tiply both bottom and top figures in each fraction 
by a number which will produce the same number 
on the bottom for all. In this case if we reduce % 
by multiplying bottom and top by 3 we have, % 5 . 
Also Yz multiplied by 5, top and bottom, gives % 5 . 
Then this gives us %5+%5-f%5=%5=/ ; 5 (reduced 
by dividing by 3). 

When it becomes necessary to add fractions 
where it is difficult to see how they can all be re- 
duced to like terms, the following process is used: 

38 



Example, add % 6 , % and % 5 . 

These must be reduced to like terms and as it is 
hard to see by what numbers to multiply, we find 
a number common to all those below the lines. 
This number is found by dividing first any one of 
these by any one other. Thus, 8|16. If the third 

2 
number is evenly divisible by the result, then do 
so, but as in this case 2 will not divide evenly into 
15 we get our new number by multiplying togeth- 
er 8, 2 and 15 which gives 240. This number is to 
be used below the line for all new or reduced frac- 
tions for each addition. Then to reduce the first 
fraction, 3 / 1G , divide 240 by 16 which gives 15. 
Multiply this number, 15 by the number above the 
line, 3, and, place the result above the line which 
gives 4 %4o for the first fraction. The second frac- 
tion, %, is reduced in like manner. Thus 240-^-8 
=30, and 30x3=90. Then the reduced fraction 
will be 9 %4o- The third fraction 2 / 15 is likewise re- 
duced. Thus 240^15=16 and 16x2=32. Then 
3 %40 i s "the third reduced fraction. 

Thus, *5/ 240+ 90/ 24()+ 32/ 240= 167/ 24() . 

PROCESSES 

(a) To add a whole number and a proper frac- 
tion all that is necessary is to annex the fraction 
to the whole number. Thus, 2014+% 6 =2014% 6 . 

(b) To add a whole number to a mixed number, 
add the two whole numbers and annex the frac- 
tion. 



39 



Example, add 456 to llf, 456+11=467, annex 
t=467|. 

(c) To add a whole number to an improper frac- 
tion, first reduce the improper fraction to a mixed 
number and proceed as above. 

Example, add 305 and 15 / 4 . 15 / 4 =3f. Then 305+ 
3=308. Annex f=308f. 

(d) To add two mixed numbers together, first 
add the two whole numbers and then add the two 
proper fractions as above and annex the two. 

Example, add together 105f and llf. First, 
105+11=116, then f+f= 9 /i 5 + 10 /i5= l9 /io- There- 
fore, we have 116 1 % 5 . 

If, as in above, the resulting fraction is an 
improper one, reduce this to a mixed number and 
add as above. Then 19 /i 5 =l 4 /i 5 , then 116+lf 15 = 

H7 4 /l5- 

When only two fractions of different terms are 
to be added, the changing of the two in- 
to like terms is done by multiplying together the 
two numbers below the line in order to obtain the 
new bottom number for both fractions. Thus, to 
add f and f we have, 3X8=24 which is the new 
number below the line for both fractions. Then to 
find the new number to use in place of f, divide 24 
by 8. Thus, 24-^-8=3. Multiply the old number 
above the line by this number and we have, 3X3 
=9. Then the new number for our reduced frac- 
tion is % 4 . In like manner reduce the f to a frac- 
tion having 24 below the line. Thus, 24-^-3=8. 
Then 2X8=16. This gives for our new fraction 

X 24. 



16 / 2 



40 



Then % 4 + 1 % 4 = 2 %4=l/<>4 (reduce to mixed 
number by dividing 25 by 24). 

CARD ROOM PROBLEMS 

1. A roll is 1 % 6 inches in diameter. It is de- 
sired to increase its diameter £ of an inch, what 
will be the resulting diameter? 

2. A man works 8f hours one day, 1\ the next 
and 10| the next. What is the total time? 

3. The weight of three samples of roving is 
56^,. 55f, and 54% grains. What is the total 
weight ? 

4. The circumference of a certain doffer is 28% 
inches. How many inches of clothing strips will be 
required to make two wraps round? 

5. 6% inches is the circumference of a certain . 
feed roll on card. How many inches of lap are 
fed when the roll turns two times? 

6. Laps are weighed as follows : 46f pounds, 
46^ pounds, 46| pounds and 46f pounds. What is 
the total weight? 

SPINNING ROOM PROBLEMS 

1. A front roll on spinning frame is 3% inches 
in circumference. How many inches of yarn will it 
deliver in 2 turns? 

3^+3%= inches. 

2. Three weighings of a yarn are made as fol- 
lows : 32f grains, 31f grains, 32^ grains. What 
is the total weight? 

3. A bobbin is 3^ inches in circumference. How 
many inches of yarn will be taken up for 3 wraps 
round ? 



41 



3|+3^+ 3i= inches. 

4. The circumference of a warper cylinder is 
37f inches. How many inches of warp is wound 
when cylinder turns 3 times? 

37f+37f+37f= inches. 

WEAVE ROOM PROBLEMS 

1. The front harness of a certain loom lifts 4f 
inches. If the back harness is to lift % inches high- 
er, what is its lift if 

4%+%= inches. 

2. 3 cuts of cloth weigh 14f, 15-J and 14^ pounds. 
What is the total weight? 

3. A man works 8f hours one day, 1\ the next 
and 8 J the next. What is the total time? 

4. The drag roll on a slasher measures 28f 
inches round. How many inches of warp are de- 
livered when roll turns twice? 

28f-f28f= inches. 



42 



Subtraction of Fractions 



In subtracting fractions one from another it is 
necessary first to reduce to like terms as in addi- 
tion. When the two fractions are in the same 
terms all that is necessary is to subtract the small- 
er number above the line from the larger one and 
place the remainder above a new line below which 
must be placed the number occurring below lines 
in the two fractions subtracted. Thus to subtract 
% from % we have : % — %=% : =44 (reduced by di- 
viding top and bottom by 2). 

Example : Subtract % 5 from |. As the two 
fractions are in different terms; that is, the two 
numbers below the line not being the same, it is 
necessary to reduce both of them to like terms. 
Thus, 60 is the only number into which both 15 
and 4 can be evenly divided so we use this as our 
new number below the line for both fractions. 
This number is usually found by simply multiply- 
ing together the two numbers below the line in 
both fractions. Then to reduce the first fraction 
% 5 divide 15 into 60. This gives 4. Multiply this 
4 by two, (the number above the line) and we have 
8. Then % is the reduced fraction. Next reduce 
| into a fraction which has 60 below the line. Thus 
60-^4=15, and 15X3, (number above) =45. Then 
4 % is the reduced fraction. 

Then, 45 / 6 o-%o= 37 /6o- 
43 



PROCESSES 

(a) To subtract a fraction from a whole num- 
ber, first subtract the fraction from 1 and reduce 
the first figure to the right in the whole number by 
1. 

Example, subtract -J from 11. 

As 1 is equal to % (because 8-^-8=1) then% — ^ 
=%. Reducing the number 11 by 1 we have 10. 
Then joining or annexing the whole number, 10, 
and the fraction % we have 10% as our remainder. 

(b) To subtract two mixed numbers, first sub- 
tract the two fractions as shown under subtraction 
of proper fractions above. Then subtract the two 
whole numbers and join the two remainders to- 
gether for the result. 

Example, subtract 2-g- from 12f or 12f — 2-J. 

First %— %=%=% (reduced by dividing both 
top and bottom by 4). Then 12— 2=10. Joining the 
two we have 10y 2 . 

Example, subtract 3% 5 from 10|. 

First reduce the two fractions to like terms. 
Then 15X4 (number below lines) =60, then 60-^- 
15=4. And 4X2=8. The first fraction then be- 
comes % . Then the next fraction f is reduced 
thus, 60-h1=15 and 15X3 (number above line) 
=45. Thus the next fraction becomes 4 % . Then 
4 %o-%o= 3 %o- Then 10-3=7. Annexing the 
whole number and the fraction we have 7 3 % as 
the resulting number. 

When the fraction in the large number is small- 
er than that in the smaller number, the figure, 1, 
must be added to the top fraction and reduced to 

44 



an improper fraction. Then 1 must be taken from 
the first right hand top figure before subtracting 
the two whole numbers. 

Example, subtract 2f from 13^ or 13^— 2f. 
As we cannot subtract % from %, we add 1 to 
% making 1%. 1%=%. Then %-%= %=V 2 (by 
dividing by 4) . Then as we added this 1 to the frac- 
tion we must reduce the first number to the right 
in the large whole number by 1. Then we have 
13 — 1=12 and 12 — 2=10. Joining the remaining 
whole number and the remaining fraction we have 

ioy 2 . 

(c)To subtract a mixed number from a whole 
number. 

First subtract the fraction from 1 as shown in 
(a) then reduce the first number to the right in the 
large number by 1. Then subtract the remaining 
two whole numbers and annex the remaining frac- 
tion and the whole number. 

Example, subtract 3% from 254 or 254—3%. As 
1=% then %- %=%. 254-1=253. Then 253- 
3=250. Result is 250%. 

CARD ROOM PROBLEMS 

1. A roll is 1 1:l /{q inches in diameter. If it is 
reduced % of an inch, what is the remaining diam- 
eter? 

%=%6- Thenl 1: ^ 6 — % 6 = 1% 6 inches diame- 
ter. 

2. A full bobbin of roving weighs 47f ounces. 
If the empty bobbin weighs 4^ ounces, what is the 
weight of roving on bobbin? 

45 



3. A roll delivers 267f inches of sliver in a min- 
ute. Another delivers 315^ inches in same time., 
what is the difference? 

4. A card delivers 147% pounds one day and 
152% pounds the next, what is the difference? 

5. A yard of laps fed to a picker weighs 3| 
pounds. If this delivers 3-§ pounds, how much 
waste is made? 

6. A lap fed to a card weighs | of a pound. If 
it delivers 90 yards of sliver weighing f of a pound, 
how much is lost in waste? 

SPINNING ROOM PROBLEMS 

1. A bobbin of yarn weighs 4| ounces. If the 
empty bobbin weighs 1^ ounces, what is the weight 
of yarn on bobbin? 

2. A warper beam when full weighs 516| 
pounds. The empty beam weighs 98^ pounds, 
what is the weight of yarn on beam? 

3. A spindle makes 8,000 turns per minute. If 
the traveler has to lag behind the spindle 216| 
turns per minute to wind on the yarn, what is the 
speed of the traveler? 

4. A certain yarn has 22% turns per inch twist. 
Another has 18% turns per inch, what is the differ- 
ence? 

5. At the beginning of an empty bobbin the 
traverse is 6 inches. If the taper is 1% inches at 
both ends, what is the length of traverse when bob- 
bin is full? 

6. A spool of full yarn weighs 18^ ounces. . If 
empty spool weighs 4| ounces, what is weight of 
yarn on spool? 



46 



WEAVE ROOM PROBLEMS 

1. The weight of warp in a cut of cloth is 9f 
pounds. How much filling is in 1 cut if 1 cut 
weighs 15^ pounds? 

2. The reed for a certain warp is 44f inches 
long. If the spread of warp in reed is only 38^ 
inches, what space should be left at each end of 
roed? 

3. If a cut of cloth is to weigh 12f pounds and 
it weighs only 1T§ pounds, how much light is it? 

4. If the opening of shed at reed is 2f inches 
when lay is back, what is the height of shuttle if 
■J clearance is allowed? 

2f — J= inches, height of shuttle. 



47 



Multiplication of Fractions 

When two proper fractions are multiplied to- 
gether the resulting number is of less value than 
either of the two fractions. In multiplying whole 
numbers the resulting number is always greater 
than either of the two. This difference must be 
kept in mind in multiplying fractions. For instance 
when we multiply 2X2 the result is 4 which is 
greather than either of the other two numbers, but 
when we multiply i by ^ the result is 5 which is 
smaller than either of the two fractions multiplied. 

PROCESSES 

(a) To multiply two proper fractions, multiply 
together the two numbers above the line and place 
the result above a new line. Then multiply together 
the two numbers below the line and place this under 
the new line. 

Example : Multiply % and % or % X V2 or read 
Y 2 of %. 3X1=3, and 2x4=8. Then the result is 

%■ 

In multiplying fractions which have large num- 
bers above and below the line the result may be ob- 
tained by using cancellation. 

Example : Multiply i% 6 by i 6 / 24 or io/ 36 X 16 / 24 . 
This may be expressed 10 X 16 



Then cancel, 







36 X24 




5 


X 


2 


5 



00 X % 3 X 9 27 

. 9 

48 



(b) To multiply a fraction by a whole number, 
multiply the number above the line by the whole 
number. Use this result for the new number above 
the line, and place the old number below the line 
under this new one. 

Example : Multiply Hie by 12 or 1:L / 16 X 12. 
11X12=132 ; then n/ 16 xl2= 132 / 16 or 8% 6 or 8%. 

(c) To multiply a whole number by a fraction, 
multiply the whole number by the number above 
the line and divide this result by the number below 
the line. 

Example : Multiply 204 by f or 204Xt or f of 204. 
204 X 3 = 612 ; then 612 -*- 5 = 122%. 

In multiplying a whole number by a fraction in 
which the number above the line is 1, it is only nec- 
essary to divide the whole number by the number 
below the line in the fraction. 

Example : Multiply 315 by | or 315 X* or £ of 315. 
Then 315 -=- 8 = 39f . 

Examples : How much is > 4 of 100 or 100 X l A ? 
How much is Y 8 of 56 or 56 X H- 
How much is Ho of 200 or 200 X Yio * 

(d) To multiply a whole number by a mixed 
number, multiply the two whole numbers first. Then 
multiply the whole number by the fraction and add 
the two results. 

Example : Multiply 272 by 4f . 

First, 272 X 4 = 1088 ; 



49 



3 272 X 3 816 
Then 272 X — = = = 102 

8 8 8 

and 1088 + 102 = 1190. 

Examples: Multiply 116 by 8%. 
Multiply 212 by 3%. 
Multiply 419 by 7%. 

(e) To multiply three or more fractions to- 
gether, place all the numbers above the lines in a 
row with the sign of multiplication between them. 
Below these draw a line and place under this all 
numbers, below the lines of -fraction, with the sign 
between them. Then to solve proceed as in can- 
cellation : 

Example : Multiply together 

10 /i 6 , %o, 15 / 3 o, %2 and i%. 

8 
Then J0 X £ XWXfi X W 1 



ifi xwxfflxffix t i6 

2^28 

This can be more clearly understood when "Can- 
cellation" is reached. 

(f) To multiply two mixed numbers together, 
reduce the two numbers to improper fractions and 
proceed as in multiplication of proper fractions. 

Example : Multiply lOf by 4f or 10| X 4f . 
10t = 32/ 3and 4 f = 23 /5 . 

32 23 736 

Then — X— = = 49 1 / 15 . 

3 5 15 

50 



CARD ROOM PROBLEMS 

1. If the circumference of a roll is 6| inches, how 
many inches of sliver will it deliver in turning 5^ 
times ? 

2. If a certain roving has 1-J turns per inch, how 
many turns would there be in 367f inches? 

3. If a roll delivers 450^ inches in a minute and 
is putting in 1-J turns per inch in roving, what is the 
spindle speed? As the spindle must make 1^ turns 
for every inch delivered, it must make 1^ X 450J for 
this delivery. 

4. A certain card produces 147f pounds per day. 
How much would it make in 6 days ? 

5. If a yard of No. 1 roving weighs 8% grains, 
how many grains would there be in 12 yards of No. 
1 roving? 

SPINNING ROOM PROBLEMS 

1. If the cylinder on a warper is 37f inches in 
circumference, how many inches of w T arp will it 
wind in turning 12-J turns? 

2. If a roll delivers 387-J inches in a minute, how 
many inches will it deliver in 60 minutes? 

3. If a man works lOf hours per day, how much 
time can he put in in 6£ days ? 

4. If 1 yard of No. 1 yarn weighs 8% grains, how 
many grains would there be in 120 yards of No. 1 
roving? 

WEAVE ROOM PROBLEMS 

1. If 52| yards of No. 1 yarn weigh 1 ounce how 
many yards of No. 1 would there be in 10^ ounces ? 

2. The width of spread in reed of certain warp 



51 



is 38f inches. How many inches of filling are used 
Avhen shuttle goes across 12 times ? 

3. How many ends would there be in warp for 
cloth 36^ inches wide and 64 ends per inch? 

4. If a sand roll on a loom is 14| inches in cir- 
cumference (around), how many inches of cloth 
would be made when it makes 3| turns? 

5. If the circumference of drag roll on slasher is 
28f inches and turns 20^ turns per minute, how 
many inches of warp are delivered in 1 minute? 



52 



Division of Fradions 



Since a fraction is an indicated form of division, 

1 
the number 1 may thus be shown, — ; because 1 -=- 1 

2 * 
= 1, and 2 may be shown — , etc. If we wish to 

1 
divide the number 1 into two equal parts, these two 

112 1 

parts will be ^ and ^ ; because 1 = — = — ■ 

2 2 2 1 
as we have seen. If then we divide 1 into two parts 

1 12 1 
and get — , then we can say - — divided by — = — , 

2 1 12 
12 1 

or ; = — . From this we see that the num- 

112 
ber below the line in the result is found by multi- 
plying together the top number in the second frac- 
tion by the bottom number in the first fraction. 
Also the top number in result is found by multiply- 
ing together the top number in first fraction by the 
bottom number in second fraction. 

3 

4 



Example : Divide f by | or f -^ ^ or — 
3 1 3X2 6 



Then 



4 2 1X4 4 

6 1 

To prove this if we multiply together — and — 

4 2 

53 



the result ought to be | just as in dividing 6 by 3 
the result 2, when multiplied by 3 gives 6. 

6 16 3 

Then — X — = — = — 
4 2 8 4 
(by dividing top and bottom by 2). 

In dividing one fraction by another,, always place 
the fraction to be divided, first ; and proceed as in- 
dicated above. 

The following diagram may be useful in dividing 
fractions : 







FIG. 1. 

PROCESS 

(a) To divide a proper fraction by a whole num- 
ber, multiply the bottom number, in the fraction to 
be divided, by the whole number, place this result 
below a new line and place above this line the same 
number occurring above the line in the fraction 
divided. 

Example : Divide % by 2 or % -^ 2 or — , 



4X2 = 8, then placing this number 
below a new line and the number, 1, 
above we have -J. 



54 



3 /l6 

Example : Divide % 6 by 4 or % 6 -f- 4 or — , 

4 
16 X 4 = 64, then the result is % 4 . 

6 /l8 

Example : Divide % s by 2 or % 8 -f- 2 or - — , 

2 
6 1 
then 18 X 2 = 36. Result = — = — (by reducing). 

36 C 

If the number above the line is evenly divisible 
by the whole number, the result may be obtainec 
by dividing this top number by the whole number 
to produce the new number above the line. The 
new number below the line in this case would be 
the same as that below the line in the number 
divided. 

9 /l6 

Example : Divide % 6 by 3 or % 6 -^- 3 or , 

3 
then 9-^3 = 3. Result : % 6 . 

(b) To divide a whole number by a fraction, 
multiply the number below the line in fraction by 
the whole number and place this result above a 
line in result. The new number below the line in 
result is same as that above the line in old fraction. 

2 
Example : Divide 2 by | or 2 -=- i or — . 



4X2 = 8. Result : — = 
1 



55 



4 
Example : Divide 4 by % 6 or 4 -r- %q or . 

64 

16 X 4 = 64. Result: — = 21% (by reduction). 

3 

(e) To divide a mixed number by a whole num- 
ber,, reduce the mixed number to an improper frac- 
tion and proceed as under (a). 

25| 

Example : Divide 25| by 16 or 25f -f- 16 or 



16 
Reduce 25|, 25 X 4 = 100, and 

103 

100 + 3 = 103 ; then 254 == and 

4 

103 103 103 

=- 16 = = = 18% 4 . 

4 4X16 fl 

(d) To divide a whole number by a mixed num- 
ber, reduce the mixed number to an improper frac- 
tion and proceed as in (b). 

24 
Example : Divide 24 by 5f or 24 -=- 5| or . 

Reduce 5f, 5 X 8 = 40 and 40 + 3 
= 43 ; then 5f = 4 % and 24 ■+- 4 % = 

24 X 8 192 

= = 420/ 



43 43 



/43- 



(e) To divide one mixed number by another, 
reduce both to improper fractions and proceed as 
in dividing one fraction by another. 

56 



25| 
Example : Divide 25| by 4^ or 25| ^- 4J or ; 

25| = 103 /4 and 4£ = 33 /s; then 

103 X 8 824 

10 % -f- 33 /s =■- - - - 



4 X 33 132 

6 32 /l32 = 6% 3 . 
(f ) When two fractions are to be divided which 
have the same numbers below the lines, or can be 
reduced to have the same numbers, all that is nec- 
essary is to divide the two numbers above the lines. 

16 /s 
Example: Divide x % by % or 16 / 8 -j- % or ; 

% ' 

then 16 -j- 4 = 4, answer. 

This is true because in dividing as shown in the 

8X16 

proceeding we have 1 % -=- % = = 4. The & 

8X4 
above and below will always cancel. 

l 3 /l6 

Example: Divide 1% 6 by 1 or 1% 6 -=- 1 or ; 



1 = i% 6 and 1% 6 = i% 6 , then i% 6 

-5" 16 /l6 = 19 /l6 = l 3 /l6 

1 

Example : Divide 1 by 1% 6 or 1 -i- 1% 6 or 



■716 



1 = i% 6 and 1% 6 = i% 6 , then i% 6 



19 /i6 = 16 /i9. 



57 



2f 
Example : Divide 2f by I5 or 2f -f- 1| or 



n 

2f = 1% and li = % = 1%, then 

' 19 /s^ 10 /8== 19 /lO = l 9 /lO. 

CARD ROOM PROBLEMS 

1. If a roll delivers 457f inches of roving in a 
minute and has a circumference of 4^ inches, how 
many turns is it making in 1 minute ? 

457f -=- 4^ == turns per' minute. 
4(457) +3 1831 



4A = 



4 4 

2X4+1 9 18 



then issi/4 -=- i8/ 4 = 1831 /is — 101 13 /i 8 turns 
per minute. 

2. If a roll delivers 358% inches of roving in a 
minute and turns 110 times, what is the circum- 
ference? 

358% -r- 110 = circumference. 

8X(358)+7 2871 
358% = = . 



110 = 88%, 

2871 880 2871 

then 1 = == 3 23 % 80 inches 

8 8 880 

circumference. 



58 



3. Circumference of a roll = diameter X 22 A- 
If a roll is 11%4 inches in circumference, what is 
the diameter? 

If 6 = 2 X 3, then 6-^2 = 3 or 6-^3= 2. 

Then 11%4 -z- 2 % = diameter. 

157 22 44 

11% 4 = and — = — . 

14 7 14 

157 44 157 

Then : • = = 3^ inches diameter. 

14 14 44 

4. One yard of No. 1 roving weighs 8% grains, 
how many yards would there be in 237% grains ? 

If 1 yard = 8% grains, then 237% grains would 

237% 
have yards. 

949 25 

237% = and 8% = — . 

4 3 

949 2847 

(multiply by 3) 



12 



25 100 

and — = (multiply by 4) 

3 12 

2847 100 2847 

then i = = 28 4 %oo yards. 

12 12 100 

5. One card produces 142% pounds of sliver per 
day. How much time would it be in making 12% 
pounds 1 



59 



713 61 

142% = and 12% = — , 

5 5 

61 713 61 

then ■- ■ = of a day. 

5 5 713 

SPINNING ROOM PROBLEMS 

1. If one spindle will produce 1^ pounds of 255 
warp in a week, how many spindles would be re- 
quired to make 8,456 pounds of the same number? 

8546 -=- If 

2. A roll delivers 327f inches of yarn in a min- 
ute. How many minutes would be required to de- 
liver 1162% inches? 

1162% -v- 327 %. 

3. There are 52^ yards of No. 1 yarn in 1 ounce. 
How many ounces of No. 1 would there be in 
1152f yards? 

4. Front roll on a certain spinning frame deliv- 
ers 382f inches of yarn per minute. If the spindle 
is running 8500 R. P. M., how many turns per inch 
would be put in? . 

8500 -=- 382f. 

WEAVE ROOM PROBLEMS 

1. If the sand roll on a loom is 14f inches in cir- 
cumference, how many times does it turn in deliv- 
ering 436 inches of cloth ? 

2. The sand roll on a certain loom is 14f inches 
in circumference. If the loom makes 684 picks while 



60 



the sand roll turns 1 time, how many picks per inch 
are put in? 

3. One loom will make 37| yards of a certain 
cloth per day. How many looms must be put on to 
make 2460 yards in a day? 

4. There are 52^ yards of No. 1 yarn in 1 ounce. 
If we have 2642f yards of yarn and it weighs 2 
ounces, what is the number ? 

521 X 2 = 105 yards of No. 1 in 2 ounces, 
then 2462f -*- 105 = No. yarn. 



61 



Decimals 

A decimal is a simple form of indicating frac- 
tions, when they are expressed in "tenths," "hun- 
dredths," "thousandths," etc. Thus Y 10 is expressed 
decimally .1. % o is expressed decimally .01. 
% 000 is expressed decimally .001. 

Therefore,, when the fraction is in "tenths" it is 
expressed decimally by having only one figure to the 
right of the decimal point. 

Examples : % = .1, 2 / 10 = .2, % = .3, 4 / 10 - .4, 

5 /io = -5, 6 /io = -6, 7 /io = -7, 8 /io = -8, % = -^ 
io/ 10 =l=1.0, ii/ 10 -iy 10 =l.l, i% - iy 10 - 1.2, etc. 

When the fraction is in "hundredths" it is ex- 
pressed decimally by having two figures to the right 
of the decimal point. 



Examples : 


Yioo = 


.01, 


t4oo = -02, 


3 /ioo 


= 


.03, 


4 /ioo 


= .04, 


%00 


= .05, 


6 /ioo = 


= -06,. %oo = 


= .07, 


8 /ioc 


• = 


.08, 


9 /ioo = 


= .09 


, 10 /ioo 


= .10, iy 100 = • 


11, 20 /ioo 


= 


.20, 


3 %>o = 


= .30, 


40 /ioo = 


= .40 


, 50 /ioo = -5C 


», 60 /ioo - 


.60, 



7 %oo - -™, so /ioo = -80, 9 %oo - -90, ™% o = 1 == 
1.00, ioy 100 = iy 100 = i.oi, io2 /loo = i2 /l00 = i.o2, 

110 /ioo = l^/ioo = 1-10, "Koo = l u /ioo = 1.U, etc. 

When the fraction is in "thousandths it is ex- 
pressed decimally by having three figures to the 
right of the decimal point. 



62 



Examples : % oo = 001, %ooo = -002, % 000 = 
-003, io/iooo - -010 or i% 000 - Y 10 o = .01, "/iooo = 
•Oil, 12 /iooo = -012, 2 %ooo = -020, ^/ 1000 = .021, 
100 /iooo = -100 or y 10 = .1, 101 /iooo = .101, ^% oo = 
.110 or ^/iooo^/ioo or .11, H&0-.1U, 120 /iooo= 
.120 or i.20 /l000 = i2 /l00 = 12 , etc. 

From the above examples we see that a cipher or 
the figure, 0, when placed to the right of the last 
figure in the decimal does not change its value. 
Hence these ciphers, or noughts, may be added or 
taken away at will. 

REDUCING FRACTIONS TO DECIMALS 

A decimal being a fraction expressed in y > 
^4 00 , etc., it is possible to change any fraction to a 
decimal. If we have the fraction % and desire to 
change it into its equivalent in % or "tenths," we 
simply divide the number, % into Yo parts. 

Then % divided by Y 10 = % -*- % - i0 A = 2% = 

2y 2 . 

Then as 2% is the result of dividing % into % 

2.y 2 

parts this number is the same as . Now as dec- 

10 
imals are shown, by placing the point,. ( . ), before 
the number instead of using 10 below the line, this 

2% 

number may be written .2y 2 . Again, if we de- 

10 
sire to. change the number ->4 ni to Yioo parts or 
""hundredths," we divide it into % 00 parts. 

Then % -h y 10 o = 10 % = 25. Therefore, if 25 is 



63 



the result of dividing % into y 100 or "hundredth" 
part, this number, 25, is the same as 2 % o- There- 
fore this number may be expressed decimally thus 
.25. If it is desired to change or reduce the frac- 
tion % to a decimal and as we do not know whether 
the resulting decimal will be in one figure, two 
figures or three figures to the right of the point, it 
is best to reduce by dividing it into "thousandths" 
or three places to the right of the point. 
Then to reduce % we have : 

% -+- Kooo = 300 % = -750. 
This is read 75 %ooo> or by reducing, 7 % o- 
From this we see that to reduce a fraction to a 
decimal we divide by the fraction %ooo- This is 
the same as multiplying the number above the line 
by 1000 and dividing into this result the number be- 
low the line. In deciding where the decimal point 
is to come in the result, it is only necessary to de- 
cide, after dividing, how many of the ciphers or 
noughts were used in the division. If only one " " 
is used, then the point would be placed so that one 
figure is to its right. If two "0's" are used, two 
figures must be to the right of the point. If three 
"0's" are used, three figures must be to the right of 
the point. 

CARD ROOM PROBLEMS 

1. If the licker-in is 9% inches in diameter ex- 
press it in decimals. As this is a mixed number it 
is only necessary to reduce the Y 2 to a decimal. 

1 X 1-000 
Then = .5, then 9% = 9.5. 



2 



64 



2. What is the decimal equivalent of a 2% inch 
feed roll on card? 

3. What is a 1% 6 inch roll equal to in decimals? 
Al 3 / 16 roll? 

4. What is the decimal equivalent of a 1% inch 
back roll on driving frame? 1% inch front roll? 

SPINNING ROOM PROBLEMS 

1. If whirl on spindle is % inch in diameter ex- 
press its diameter in decimals. 7 X 1000 = 7000, 
then 7000 -f- 8 = .875 inches. 

2. If whirl is 1% 6 inches, express it in decimal. 
As this is a mixed number it is only necessary to re- 
duce the fraction and join the two. Then 1 X 1000 
= 1000,. and 1000 -=- 16 = .062+. As this does not 
come out even in dividing we add another "0" and 
obtain .0625. Then result = 1.0625. 

3. If front roll on spinning frame is V/ 8 inches in 
diameter, express it in decimals. 

4. If a certain bobbin contains 3% ounces of 
yarn, express it in decimals. 

WEAVE ROOM PROBLEMS 

1. 1 yard of No. 1 yarn weighs 8% grains; ex- 
press it in decimals. 

1 x 1000 = 1000 and 1000 -4- 3 = .333+ 
then we have 8.333. 

2. If a reed has an average of 30% dents per 
inch, express it in decimals. 

3 X 1000 = 3000, and 3000 -h 4 = .750 
Then 30% = 30.75. 

3. A certain bobbin has 3% ounces of yarn in it ; 
express it in decimals. 



65 



4. The spread in reed for a certain cloth is 38% 
inches ; express it in decimals. 

UNITED STATES MONEY 

The money of our Government is expressed in 
decimals. There are 100 cents in a dollar. Then 
one cent would be %oo of a dollar and is expressed 
.01. Twenty cents is 2 %oo of a dollar and is ex- 
pressed thus, .20. The dollar sign ($) is usually 
placed before the amount indicated. Thus fifteen 
cents is written $.15, one dollar and fifty one cents, 
$1.51, etc. Y 10 of a cent is called a mill and would 
De /looo °f a dollar and is written or expressed 
then, $.001. 

Write in figures the f ollowing : 

One dollar, fifty-three and one-half cents. $1,535. 

Twenty dollars, forty-two and one-fourth cents. 
$20.4225. 

Fifty-one dollars and fifteen cents. $51.15. 



66 



Addition of Decimals 



In finding the sum of two or more decimals, place 
them one under the other so that the decimal points 
will come directly under each other and proceed as 
in addition of whole numbers. 

Example : Add .125, .364, .7821 and .1234. 
.125 
.364 
.7821 
.1234 



1.3945 



Place the point in the result directly under all the 
points in numbers added. This will always give as 
many figures to the right of point as there are fig- 
ures in the longest decimal used. In the above 
addition the longest decimal contained four figures 
and we must have four figures to the right of the 
point in the result. 

CARD ROOM PROBLEMS 

1. If 8 laps weigh as follows : 42% pounds, 41% 
pounds, 40% pounds, 41% pounds, 42% pounds, 41% 
pounds,, 42% pounds and 41% pounds, what is the 
total weight expressed in decimals? 

% = -5, % = -75, % = .875, % = .25. 

Then we have 



67 



42.5 

41.75 

40.875 

41.25 

42.75 

41.875 

42.5 

41.75 

335.250 pounds = 335% pounds 

2. If the overseer makes $31.25 per week, sec- 
ond hand $21.43 and six fixers $19.38 each, what is 
the total wages paid? 

3. If four samples of roving are weighed as fol- 
lows, what is the total? 52.4 grains, 53.1 grains, 
52.4 grains and 52.6 grains. 

4. The circumference of a certain roll is 3.425 
inches; how many inches of roving will it deliver 
in turning 3 times? 

3.425 + 3.425 +.3.425 = inches. 

SPINNING ROOM PROBLEMS 

1. A roll is 3.1416 inches in circumference ; how 
many inches of yarn will it deliver in turning 3 
times ? 

3.1416 + 3.1416 + 3.1416= inches. 

2. The overseer gets $31.42, the second hand 
$22.60, 4 section hands $18.25 each; what are the 
total wages paid? 

3. A spinning room produces 2624.32 pounds of 
28 's, 4681.47 pounds of 30 's, 12822.67 pounds of 40 's 
and 15461.38 pounds of 22 's in a week; what is the 
total production? 

68 



4. 4 samples of the same yarn are taken and 
weigh 32.4 grains, 31.8 grains, 32.2 grains and 31.9 
grains; what is the total weight? 

WEAVE ROOM PROBLEMS 

1. A certain loom produces 38.24 yards in a day ; 
how many yards would it produce in 4 days? 

38.24 + 38.24 + 38.24 + 38.24 = yards. 

2. Four samples of cloth weigh 15.27 pounds, 
14.84 pounds, 14.92 pounds and 15.18 pounds ; what 
is the total weight? 

3. Overseer receives $32.41, two second hands 
$22.50 each and 8 section hands $18.75 ; what is the 
total wage paid? 

4. Supplies are received as follows: 2 shuttles 
$.94 each, 2 reeds $1.14 each, 4 picker sticks at $.09 
each ; what is the total cost ? 



69 



Subtra&ion of Decimals 



In subtracting one decimal from another,, place 
the one number under the other as in addition of 
decimals and proceed as in subtraction of whole 
numbers. 

Example : Subtract .1234 from .5216. 
.5216 
.1234 



.3982 

Example : Subtract 1.923 from 3.134. 

3.134 
1.923 



1.211 ' 

Example : Subtract .785 from 2. 
2.000 

.785 



1.215 
CARD ROOM PROBLEMS 

1. A yard of a certain lap weighs 5261.24 
grains. If it makes 90 yards of sliver which weigh 
5182.36 grains, how many grains are lost? 

2. Back roll on drawing frame takes in 110.26 
inches of sliver, while the front roll is delivering 
624.82 inches. How much more does front roll de- 
liver than back roll? 

70 



3. A lap is supposed to weigh 5265.64 grains. If 
it weighs 5372.72 grains, how much too heavy is it? 

4. A full bobbin of roving weighs 47.64 ounces. 
If the empty bobbin weighs 4.34 ounces, how much 
roving is on the bobbin? 

SPINNING ROOM PROBLEMS 

1. A full bobbin weighs 5.26 ounces. If the 
empty bobbin weighs 2.42 ounces, how much yarn 
is on bobbin? 

2. One roll is 3.52 inches in circumference and 
another is 4.23 inches ; how much more roving will 
the latter roll deliver than first when both rolls will 
have turned two times? 

3. A spool when full of yarn is 12.56 inches in 
circumference. When near empty it is 3.75 inches 
in circumference. How much more yarn will be 
unwound in 2 turns when full than will be unwound 
in 2 turns when near empty? 

12.56+12.56 = 25.12 inches unwound when full. 
3.75-j- 3.75 = 7.50 inches unwound when near 

[empty. 

17.62 inches more. 

4. The breaking weight of a 20 's warp is 88.34 
pounds, of a 30 's warp it is 66.28 pounds. What is 
the difference? 

WEAVE ROOM PROBLEMS 

1. A reed is 45.25 inches long, the warp is to be 
spread 39.6 inches. How many inches must be left 
on each end of reed? 

2. A cloth is to weigh 4.84 yards per pound 
which makes a cut of 60 yards weigh 12.39 pounds. 



71 



If a cut weighs 13.25 pounds, how much loss would 
this be in every cut? 

3. A certain cloth weighs 2.85 ounces per yard. 
If the warp in 1 yard weighs 1.93 ounces, what is 
the weight of the filling? 

4. A sample of sized yarn weighs 26.8 grains, 
after the size has been washed out and yarn dried 
it weighs 25.9 grains. How much size was in the 
yarn? 



72 



Multiplication of Decimals 

To multiply one decimal by another place the two 
numbers as in multiplication of whole numbers re- 
gardless .of where the two decimal points are. 
Proceed to multiply as in whole numbers and place 
the decimal point in the result so that there will 
be as many figures to its right as are counted to the 
right of the points in both numbers multiplied. 

Example : Multiply .3041 by .012. 

.3041 
.012 



6082 
3041 



36492=.0036492 

In the bottom number multiplied we count three 
figures to the right of the point, and in the top num- 
ber we count four figures. Both together this gives 
seven. Then we must have seven figures to the right 
of point in the result. As our multiplication only 
produced five figures we must make up the seven 
by prefixing "0's." Therefore we have the number 
.0036492 as shown to the right above. 

When two decimals are multiplied together the 
resulting decimal is smaller than either of the two 
multiplied. 

Example : .25 X .25=.0625. 

When a whole number is multiplied by a decimal 



73 



the resulting number is smaller than the whole 
number. 

Example: 4X .25=1.00. 

To multiply two mixed, numbers in decimals. 

Example : Multiply 2.75 by 1.785. 

1.785 
2.75 



8925 
12495 
3570 

4.90875 

CARD ROOM PROBLEMS 

1. The feed roll on a card is 2.5 inches. If it runs 
.85 revolutions in a minute, how many inches of lap 
are fed to licker-in? 

When the roll makes one turn it will deliver as 
many inches as there are inches in its circumference 
(distance around). The circumference is found by 
multiplying the diameter (distance through) by the 
standard number, 3.1416. 

Then 2.5X3.1416=7.854 inches delivered in 1 turn. 

If it makes .85 in 1 minute, we have, 

7. 854X .85=6.6759 inches delivered to licker-in 
in one minute. 

2. The doffer on a certain card is 27% = 27.75 
inches diameter through clothing. If it makes 10.3 
revolutions per minute, how many inches of sliver 
will it produce in one minute? 



74 



Circumference=27.75X 3.1416=87.179 inches ; 
then 87.179X10.3=879.94 inches delivered in 

one minute. 

3. The diameter of a metallic roll which is 1% 
inches is figured as 1.83 inches diameter. If front 
roll on a drawing frame is 1%=1.83 inches in diam- 
eter and is running 300 revolutions per minute, how 

many inches of sliver will it deliver in 1 hour? 

«» 

Circumference=1.83X 3.1416=5.749 inches 
then 5.749X300=1724.700 inches delivered in 

one minute, 
1 hour=60 minutes, then 1724.7X60=103482.0 

inches in 1 hour. 

4. A man works 24% hours at the rate of 2V-/ 2 
cents per hour, what is the total amount he should 
receive ? 

24% hours=24.75 

2iy 2 cents=$.215 

then 24.75 X-215=$5.32. 

5. If the back roll on a slubber takes in 275.24 
inches of sliver in a minute and the draft is 3.75, 
how many inches will the front roll deliver in 1 
minute ? 

Draft X inches fed = inches delivered; 
then 275.24X3.75=1032.15 inches delivered. 

6. The twist per inch for roving is found by mul- 
tiplying the square root of the hank roving to be 
made by the standard, 1.2. If 4 H. R. is to be made, 
how many turns per inch should be inserted? 

V4=2. Then 2x1.2=2.4 turns per inch. 

75 



Find twists for following numbers of H. R. : 

.25, V.25=.5, then .5 Xl.2--=.60 turns per inch. 

.75, V.~75=.S66, then .866X1.2=1.039 turns per inch. 

.2, Vl*~=1.414, then 1.414 X 1.2=1.696 turns per inch. 

SPINNING ROOM PROBLEMS 

1. The weight of yarn on a bobbin from a 2% ring 
and 7 inch traverse is about 3.875 ounces. How 
many ounces would there be in a doff from 300 
spindles ? 

3.875X300=1162.500 ounces. 

2. 1 yard of No. 1 yarn weighs 8.333 grains,, what 
would 550 yards of a No. 1 weigh? 

8.333X550=4583.15 grains. 

3. If the ratio of cylinder to whirl is 1 to 7.25, 
what would be the speed of spindle if cylinder runs 
960 revolutions per minute? 

This ratio means that when cylinder makes 1 turn 
the whirl and spindle makes 7.25 turns. Then the 
whirl turns 7.25 times as fast as the cylinder and if 
the cylinder makes 960 turns the spindle will make, 
960X7.25=6960.00 revolutions per minute. 

4. The twist per inch for warp yarn is found by 
multiplying the standard number, 4.75, by the square 
root of the number of yarn. Find the turns per 
inch to be put in following warp numbers : 

36, V36=6, then 4.75x6=28.50 turns per inch. 
28, V28=5.291, then 4.75x5.291=25.13 turns 

inch. 
30, V-30=5.477, then 4.75x5.477 = 26.01 turns 



inch. 

76 



per 
per 



5. The twist per inch for filling is found by mul- 
tiplying the standard number, 3.50 by the square 
root of the number of yarn. 

Find the turns per inch to be put into the follow- 
ing numbers : 

38, V38"= 6.164, then 3.50 X 6.164 = 21.57 turns 
per inch. 

40, ViO = 6.324, then 3.50 X 6.324 = 22.13 turns 
per inch. 

44, V44 = 6.648, then 3.50 X 6.648 = 23.26 turns 
per inch. 

6. The diameter of a front roll on a certain spin- 
ning frame is 1%=1.125 inches, and it is running 
112.8 revolutions per minute. How many inches 
will it deliver in 1 minute? 

In one turn the roll will deliver as many inches 
as there are inches in its circumference (distance 
around). The circumference is found by multiply- 
ing the diameter by 3.1416. Then 1.125 X 3.1416 
= 3.5343 circumference = inches delivered in one 
revolution. Then if it turns 112.8 times in one min- 
ute, the number of inches delivered in 1 minute will 
be 3.5343 X 112.8 = 398.669 inches delivered in 1 
minute. 

7. If a girl runs 10 sides and is paid 17 cents 
per day per side, what should be her pay for 6 days' 
work? 

17 cents = $.17. Then 10 X $-17 = $1.70 
per day, and $1.70 X 6 = $10.20 pay for 6 days. 

WEAVE ROOM PROBLEMS 

1. If a reed has 28.75 dents on 1 inch, how many 
dents would there be on 37.84 inches ? 



77 



28.75 X 37.84 = 1087.9 dents. . 

2. A cloth has an average of 84.72 ends in 1 
inch. If it is to be 36% inches wide, how many ends 
should there be in the warp? 

36y 2 = 36.5. Then 84.72 X 36.5 = 3092.28 ends ; 
this would be called 3092 ends. 

3. The sand roll on a certain loom is 14.75 inches 
around. How many inches of cloth will be taken up 
when the roll turns 5.375 turns? 

14.75 X 5.375 = 79.28 inches. 

4. If the diameter of drag roll on slasher is 9.452 
inches and it is turning 19.52 revolutions per min- 
ute, how many inches of warp will be delivered in 
1 hour? 

When the roll makes one turn it will deliver as 
many inches of yarn as there are inches around the 
roll. The number of inches around the roll (cir- 
cumference) is found by multiplying the diameter 
by 3.1416. 

Then 9.452 X 3.1416 = 29.694 inches delivered in 
1 turn. 

And 29.694 X 19.52 = 579.626 inches delivered in 
1 minute. 

Then 579.626 X 60 (minutes in 1 hour) = 
34777.560 inches delivered in 1 hour. 

5. If "cloth is to weigh 5.64 yards per pound,, how 
many yards would there be in 16.75 pounds? 

1 pound = 5.64 yards. Then 16.75 pounds 
would have 16.75 X 5.64 = 94.47 yards. 

6. If a weave room prod es in a week, 5236 



78 



pounds of a cloth which weighs 4.75 yards per 
pound, how many yards would that be? 

5236 X 4.75 = 24871 yards. 

7. There are 52.5 yards of No. 1 yarn in 1 ounce. 
How many yards would there be in 14.62 ounces 
of No. 1?" 

14.6 X 52.5 = 766.5 yards. 

8. If there are 52.5 yards in 1 ounce of No. 1 
yarn, how many yards would there be in 1 ounce 
of No. 30 yarn! 

52.5 X 30 = 1575.0 yards. 

9. The spread in reed of a certain warp is 39.84 
inches. How many inches of filling would be put 
into 1 inch of cloth if there are 64 picks to the inch? 

39.84 X 64 = 2549.76 inches. 

10. If a weaver receives 24 cents a cut and pro- 
duces 64 cuts in one week, what should be his pay? 

24 cents = $.24. 
Then 64 X $.24 = $15.36 per week. 



79 



Division of Decimals 



In dividing one decimal by another proceed as in 
division of whole numbers. In placing the decimal 
point in the result, place it so that there will be as 
many figures to its right as the difference between 
the numbers to its right in the two numbers divided. 

Example : Divide .4562 by .0136. 

.0136 ) .4562000 ( 33.544 
408 

482 
408 

740 

680 

600 
544 

560 
544 

In the above example we place the numbers 
.0136 and .4562 as shown and divide as in whole 
numbers disregarding the decimal points and any 
ciphers or "0's" which occur between the point and 
the first figure in the number. In this case, then we 
divide 136 into 456 and it goes 3 times with 48 over 
as seen. Drawing down the next figure, 2, we have 
482 into which we again divide 136. After the 
figure, 2, is brought down We see that the division 
is not complete, because we have 74 left over. Then 
we add "ciphers" or "noughts" to the number di- 



80 



vided bringing these down in the process as if they 
were figures. To decide where the point is to come 
in the result, keep in mind the number of figures to 
right of the point in number divided by. When we 
have used up as many figures to the right of point in 
number divided as there are figures to the right in 
the number divided by, the point should be placed to 
the right of the figure in result. In this example, 
after we have drawn down the figure, 2, we see that 
we have used as many figures to the right as we 
have in the number .0136. Then after making the 
division, "136 into 482 = 3," place the point after 
this figure, 3. As many "ciphers" or "noughts" 
may then be added to the number, .4562,. accord- 
ing as to how many figures Ave desire to the right 
of point in result. 

Example : Divide .0231 by .001. 

A 

.001) .023 11 (23.1 

9 



The line, A, may be drawn between two figures to 
show where the point is to go in the result. We 
place this line between 3 and 1 thus cutting off three 
figures to the right of the point in the number .0231, 
to equal the total number of figures to the right of 
point in number, .001. Thus after we have brought 
down the figure, 3, and divided 1 into it to produce 
3 we place the point in the result after this figure, 3. 

Example: Divide 2.0534 by .025. 



81 



.025 ) 2.053J40 ( 82.136. 
2 00 

53 
50 

34 
25 

90 
75 

150 
150 

This division can be proved correct by multiply- 
ing the result 82.136 by the number divided by, 
.025.. 

Thus, 82.136 X -025 = 2.053400, number divided. 

Example : Divide 1 by .001. 

.001 ) 1.0001 ( 1000. 



000 



In this case it was necessary to add as many 
"noughts" to the figure, 1, as we have in the figure 
.001. Remember always that when "noughts" are 
added to a whole number, as in this case, the point 
must be placed after the whole number. 



Example : Divide 53.5 by 2.352. 

2.352 ) 53.500|00 ( 22.74 
47 04 

6 460 
4 704 

17560 
16464 

10960 
9408 



Example : Divide .123 by 100. 

100. ) |.12300 ( .00123 
100 

230 

200 

300 
300 



CARD ROOM PROBLEMS 

1. A lap weighs 40.75 pounds and is 56 yards 
long; how many ounces does 1 yard weigh? 

1 pound = 16 ounces ; 

therefore, 40.75 X 16 = 652 ounces, wt, of 56 yards ; 

then 652 -4- 56 = 11.6+ ounces per yard. 

2. Laps are being made weighing 40.75 pounds. 
If 30 bales of cotton averaging 475 pounds per bale 
are opened, how many laps will it make? 

30 X 475 == 14250 pounds of cotton run ; 
then 14250 --=- 40.75 = 349.6 laps made. 

3. A lap 56 yards long weighs 41.25 pounds ; how 



many yards of sliver will be made weighing 54.5 
grains per yard? 

1 pound = 7000 grains; 
then 41.25 pounds = 41.25 X 7000 = 288750 grains ; 
each yard of sliver weighs 54.5 grains,, 
288750 

then = 5298.1 yards of sliver. 

54.5 

4. A feed roll on a card is 2% inches in diameter ; 
how many times will it turn in taking in 1 yard of 
lap? 

2.5 X 3.1416 = 7.854" circumference; 

1 yard = 36 inches ; 

then 36 -f- 7.854" = 4.583 turns. 

5. The cylinder on a card is 50" in diameter and 
is 45 inches wide. How many feet of card fillet will 
it take to cover it, if fillet is 2" wide ? 

3.1416 X 50 = 157.08 inches circumference ; 
so it will require 157.08 inches of fillet for 
each round. The taper will be 2" in each 
157.08 inches. 45 -=- 2 = 22% rounds of fillet 
to cover 45 inches; then 157.08 X 22% = 
3534.3 inches of fillet to cover cylinder. Or 
3534.3 -r- 12 = 294.5 lineal feet of fillet. 

6. A man draws $12.75 for running frames at 8 
cents per hank. How many hanks did he run? 

7. Fine frame spindles are running 1200 R.P.M. 
and front roll is delivering 629.42 inches per minute ; 
how many turns per inch are being put in? 

SPINNING ROOM PROBLEMS 

1. Spool when full is 12.56 inches in circumfer- 
ence. When near empty it is 3.75 inches in circum- 

84 



ference. How many more turns must the latter size 
spool make to deliver the same amount of yarn as 
the full spool? 

In 1 turn of the full spool it will deliver 12.56 
inches of yarn, and in 1 turn of the empty spool it 
will deliver 3.75 inches; therefore, 12.56 -=- 3.75 = 
3.35 times as fast. 

Example : If the spool above when full turns 100 
R.P.M. to deliver a certain amount of yarn it will 
have to turn 100 X 3.35 = 335 R.P.M. when near 
empty to deliver the same amount. Why is the 
cone drive put on a warper? 

2. If spindle is running 8400 R.P.M. and front 
roll is delivering 437.48 inches per minute, how 
many turns per inch are being put in? 

8400.00 -f- 437.48 = 19.2. turns per inch. 

3. If spindle speed is 8600 R.P.M. on spinning 
frame and yarn No. is 36 warp, what is the front 
roll delivery in yards in 1 hour? 

V 36 = 6 and 4.75 X 6 = 28.5 turns per inch, 

then 8600^-28.5 = 301.75 inches delivery in 1 minute 
and 301.75 X 60 = 18105 inches delivery in 1 hour, 
then 18105 ~ 36 (inches per yd.) = 502.91 yards 
delivery in 1 hour. 

4. If the front roll is 1 inch in diameter, how 
many turns per minute is it making for the deliv- 
ery in above example? 

It is delivering 301.75 inches per minute ; in 1 
turn it will deliver 1 X 3.1416 = 3.1416 inches (be- 
cause its circumference is 3.1416")- Then 301.75 
-r- 3.1416 = 96.4 revolutions per minute. 



85 



WEAVE ROOM PROBLEMS 

1. If cloth weighs 5.64 yards per pound, how 
many ounces would 1 yard weigh? 

5.64 yards = 1 pound = 16 ounces ; 
then 16 -=- 5.64 = 2.83 ounces weight of 1 yard. 

2. If cloth weighs 4.75 yards per pound, what 
would be the weight of a 60-yard cut? 

60 -f- 4.75 = 12.63 pounds. 

3. 6 repeats of a fancy pattern measure 7.25 
inches. Each repeat has 96 ends. What is the 
"over-all" construction or average ends per inch? 

96 ends X 6 repeats = 576 ends on 7.25 inches ; 
then 576 -f- 7.25 = 79.44 average ends per inch. 

4. If a sand roll on loom is 14.35 inches in cir- 
cumference, how many times will it turn in deliv- 
ering 1 yard of cloth? 

1 yard = 36 inches; 
then 36 -=- 14.35 = 2.5 turns. 



86 



Ratio and Proportion 

The relation of one quantity to another of the 
same kind is called the ratio of the one to the 
ifcher. Thus if Ave have two pulleys one running 
10€ turns per minute and the other 200 turns in the 
same time the first one will make 1 turn while the 
second is making 2 and we say that the ratio of the 
first to the second is as 1 to 2. Also if we have a 
pulley or drum which in turning one time drives by 
a band or belt another one which makes 6 turns in 
the same time, the ratio of speed between the two is 
1 to 6. The ratio between numbers is usually in- 
dicated by the sign, :, and is read "is to." Thus 
in the latter example above Ave may express the 
ratio thus, 1 : 6. This is read "1 is to 6." 

In some cases the ratio betAveen numbers is indi- 
cated by placing one above a. line and the other be- 
low. Thus in shoAving the ratio betAveen tAvo num- 
bers say 1 and 5 Ave may use the 1 above the line 

1 
and the 5 beloAv, thus — . This is read also "1 is to 5. " 
5 

When Ave have two ratios which are equal to each 
other this is said to be a proportion. Thus if Ave 
have tAvo pulleys one running 100 turns per minute 
and the other 200 turns and compare these speeds 
to tAvo more pulleys one of Avhich makes 300 Avhile 
the other makes 600 turns, these tAvo ratios are said 
to be in proportion. This is true because the ratio 
of the first tAvo pulleys is 1 : 2 Avhile the second two 



87 



also have the ratio 1:2. Then these two relative 
speeds are in proportion and we may say "100 : 200 
= (equals) 300 : 600." In writing proportions it 
is sometimes customary to use, instead of the equal- 
ity sign (=), the sign, : :, which is read "as." Thus 
in the above proportion it may be written thus : 
100 : 200 : : 300 : 600, and is read "100 is to 200 as 
300 is to 600." 

In the above proportion we can readily see that 
the second term (200) in the first ratio when mul- 
tiplied by the first term, 300,. in the second ratio is 
equal to the number obtained by multiplying the 
other two terms together. Thus 200 X 300 = 60000, 
and 100 X 600 = 60000. This must be true in all 
proportions. Therefore in any proportion the two 
inside terms multiplied together are equal to the 
two outside terms multiplied together. From this 
law it is plain that if three terms are given the 
fourth or remaining term can be found by multi- 
plying together the two inside or outside terms as 
the case may be, and dividing this result by the 
third term given. 

Thus if we again use the proportion 100 : 200 : : 
300 : 600, any one of the terms may be found, if not 
given, if the other three are known. The term 
which is not given is usually represented by the 
letter X. Thus if we leave out the third term, 300, 
and insert the letter, X, we will have, 100 : 200 : : 

100 X 600 

X : 600. Then to solve we have, = 300, 

200 

which is the unknown number, X, or the third term, 
300. In like manner any one of the other terms 



can be found if the other three are given. Thus, 

200 X 300 

X : 200 : : 300 : 600. Then = 100 or X, 

600 

or the first term. 

100 X 600 

Also, 100 : X : : 300 : 600. Then = 

300 

200 or X or the second term. And 100 : 200 : : 

200 X 300 

300 : X. Then = 600 or X or the fourth 

100 

term. This method is commonly called the "Rule 
of Three." Proportions are usually stated in con- 
sidering the relative values of cause and effect. 
Thus if a man in 6 days makes 12 dollars he will 
make in 12 days 24 dollars. In this case he works 
6 days and makes 12 dollars. The ratio of time to 
money or work to pay is 6 : 12 or 1 : 2. As he 
works at the same rate of pay for the whole 12 days 
and receives 24 dollars the ratio is again 12 : 24 or 
1 : 2 and these ratios being equal are in proportion 
and may be stated, 6 : 12 : : 12 : 24. In this we 
see that, time : pay : : time : pay, provided the rate 
of pay per day has remained the same. We also see 
that the two inside terms, 12 and 12, when multi- 
plied together will give 144; and the two outside 
terms when multiplied together will give 144. Again 
if it is desired to find any one of these terms when 
the other three are given it is only necessary to mul- 
tiply either the two inside or outside terms together 
and divide this by the remaining term. 



89 



Example : If a man works 6 days and makes 18 
dollars, how many days will he have to work at the 
same rate of pay to make 90 dollars ? 

Proportion 6 : 18 :: X :,90. 

6 X 90 

Then = 30 days. 

18 

Example : If a man works 6 days and makes 18 
dollars, how much will he make in 60 days \ 

Proportion 6 : 18 : : 60 : X. 

18 X 60 

Then = 180 dollars. 

6 

In like manner the other two terms may be found. 
CARD ROOM PROBLEMS 

The draught gear on a card is directly propor- 
tioned to the weight in grains per yard of sliver 
made, provided the weight of the lap has not been 
changed. Thus if an 18 teeth gear makes a 54 
grain sliver, what gear will be required to make a 
60 grain sliver from same lap? 

The proportion is, 18 : 54 : : X : 60 ; because if an 
18 teeth gear produces a 54 grain sliver, X No. of 
teeth will be required to produce a 60 grain sliver. 

To solve multiply together the two outside terms, 
18 and 60, and divide this by the remaining inside 
term 54. The result is the unknown number of teeth 
or the unknown quantity, X. 

18 X 60 

Thus = 20 — teeth draught gear required. 

54 

90 



1. If a 20 teeth draught gear is making a 60 
grain sliver from a certain lap, an 18 teeth draught 
gear Avill produce what weight sliver from same 
lap? 

Proportion 20 : 60 : : 18 : X. 
Solve : 

2. The production gear on the card is also direct- 
ly proportional to the pounds production. Thus, if 
a 25 teeth production gear is making 150 pounds per 
day of a certain weight sliver, what gear will be in- 
quired to make 126 pounds per day of same sliver 1 
If a 25 teeth gear makes 150 pounds then X num- 
ber of teeth will make 126 pounds and avc have : 

25 : 150 :: X : 126 ; 

126 X 25 

and = X = 21 — teeth production gear. 

150 

3. If a 25 teeth production gear on card makes 
150 pounds of sliver in a day, what size gear would 
produce 144 pounds in a day of the same weight 
sliver ? 

State proportion and solve. 

SPINNING ROOM PROBLEMS 

1. If an 8 inch cylinder is driving a 1 inch whirl.. 
how many times would the whirl turn while the 
cylinder turns one time? Answer 8. What is the 
ratio of cylinder to whirl? 1 : 8. 

2. If ratio of cylinder to whirl is 1 : 8, how many 
turns per minute must the cylinder make to give a 
spindle speed of 6400 turns per minute? 



91 



Proportion 1 : 8 : : X : 6400 ; 

6400 
then X = = 800 turns per minute of cylinder. 

8 

3. If it takes 4 spindles to produce 1 pound of 
yarn per day, how many pounds would be made on 
30000 spindles? 

Proportion 4:1:: 30000 : X ; 

then 4 X X = 1 X 30000 ; 

30000 

then X = = 7500 pounds. 

4 

4. If an average number of sides per girl is 14, 
how many girls would be required to run 280 sides? 

Proportion 1 : 14 : : X : 280 ; 

14 X X = 1 X 280 ; 

280 

X = = 20 girls. 

14 

WEAVE ROOM PROBLEMS 

1. If a 32 teeth gear gives 64 picks per inch, 
what pick gear will give 100 picks ? 

Proportion 32 : 64 : : X : 100 ; 

64 X X = 32 X 100 ; 

32 X 100 

and X = = 50 teeth gear. 

64 

2. 8 looms produce 20 cuts in a week. How 
many looms would it take to make 250 cuts per 
week? 



92 



Proportion 8 : 20 : : X : 250. 

20 X X = 8 X 250 ; 

8X250 
X = = 100 looms. 



20 

3. If crank shaft on loom makes 2 turns while 
the cam shaft makes 1 turn, what is the ratio of 
speed of cam to crank shaft ? Answer 1 : 2. 

If the crank shaft has a gear of 50 teeth, how 
many teeth on cam shaft gear? 

1 : 2 :: 50 : X. 

1 X X = 2 X 50; 
X = 100 teeth. 

4. If 4 yards of cloth weigh 1 pound, how many 
yards of cloth would there be in 15 pounds ? 

4 : 1 : : X : 15. 

1 X X = 15 X 4; 
X = 60 yards. 

5. If 4 yards of cloth weigh 1 pound, how many 
pounds would there be in 60 yards? 

4 : 1 : : 60 : X. 

4 X X = 1 X 60; 

60 
X = — = 15 pounds. 
4 



93 



6. If a spread in reed of 38 inches makes 36 inch 
cloth, what must be the spread to make 72 inch cloth 
of the same construction? 

38 : 36 : : X : 72 
36 X X = 72 X 38 
72 X 38 

X = = 76 spread in reed. 

36 



94 



Square and Square Root 

When one number is multiplied by itself the re- 
sulting number is said to be the square of the num- 
ber thus multiplied. If we multiply 2 by itself the 
result is 4 and the number, 4, is said to be the square 
of 2. Thus to square any number all that is neces- 
sary is to multiply that number by itself. 

EXERCISE 

What is the square of 4? 4 X 4 = 16. 

What is the square of 5 •? 5 X 5 = 25. 

What is the square of 6? 6 X 6 = 36. 

What is the square of 10? 10 X 10 = 100. 

What is the square of 20 ? 20 X 20 = 400. 

What is the square of 25? 25 X 25 == 625. 

When we find the number which was used in 
multiplying by itself to produce a certain other 
number this number is said to be the square root of 
the one thus obtained. For instance, when we mul- 
tiply 2 by itself to produce its square, 4, then the 
number 2 is said to be the square root of 4. Thus, 4 
is the square root of 16 because 4 multiplied by it- 
self produces 16. In like manner 5 is the square 
root of 25, 6 the square root of 36, 8 the square root 
of 64, 10 the square root of 100 and so on. When- 
ever we extract the square root of any number this 
square root when multiplied by itself must produce 
the number from which we started. 

In finding the square root of numbers of large 

95 



value,, or those in which the square root is not a 
whole number, the following process is used : 
Find the square root of 14884. 

1,48,84 ( 122 

1 

20 48 
22 44 

240 4 84 
242 4 84 

The first thing to do is to mark off, as shown, the 
number into groups of 2 figures starting at the 
right of the number. Then select a number which 
when multiplied by itself is not greater than the 
first group on the left of number. In this case only 
one figure is in this last group on the left. This is 
the figure 1, and we select 1 as our first figure in the 
result because 1 multiplied by itself produces 1 
which is not greater than our first group. This fig- 
ure is then placed to right of line as shown and its 
square which is also 1 is placed under the first 
group to the left. This is then subtracted from the 
first group and the next group of two figures 
brought down. In this case 1 from 1 leaves noth- 
ing and our only remaining number is 48. Then 
multiply the figure, 1, in the result by figure, 2, 
and annex to it the .figure, 0. This gives 20 as 
shown to the left of 48. Then divide this number 
20, into 48. This gives 2, and we place this next to 
1 in the result. This figure 2 must next be added to 
20 giving the number 22 which is placed directly 
under 20. Then multiply 22 by 2 in result and place 
this result under 48 and subtract as in long division. 



96 



Subtracting then we obtain 4 and then drawing- 
down the next group, 84, we have 484. Again mul- 
tiply the result, which is now 12, by 2 and add to 
this the figure, 0. This gives 240 and we divide this 
into 484 and find that will give a result of 2. This 
is placed to right of 12 and is also added to 240 
making 242 which is placed below 240. Then mul- 
tiply this 242 by the figure last obtained in result, 
namely 2„ and we obtain 484 which is placed below 
the last remainder 484. The square root then of 
14884 is the whole number, 122. If it becomes nec- 
essary to find the square root of a number which 
is not a whole number, the following method is 
given. This is the same as the one just given ex- 
cept that the figures, 0, in groups of two are placed 
to the right of the whole number after the decimal 
point has been inserted. 

Example : Extract the square root of 30. 

30.00,00 ( 5.47 
25 

100 5 00 
104 4 16 

JL080 84 00 
1087 76 09 

The only thing to note in such examples is that 
the whole number is put to the left of the decimal 
point, and must be divided into groups of two fig- 
ures from right to left and the ciphers or figure 0, 
are put to the right of the decimal point and must 
be divided into groups of two from left to right. 
The following numbers are given showing how to 



97 



divide into groups so as to produce two figures to 
the right of the decimal in the result. 

1,33.00,00. .20.00,00. 2,63.00,00. 

21,53.00,00. 1,42,53.00,00. 2.00,00. 

EXAMPLES FOR CARD ROOM 

The twist per inch to be inserted into any size or 
number of hank roving is found by multiplying the 
square root of the number of roving by a certain 
given standard number. 

Run the following to two figures to the right of 
decimal point. 

Find the square root of No. 2 hank roving. 

Find the square root of No. 3 hank roving. 

The sign of square root is made thus, ^J and 
when placed before a number it means that this num- 
ber is to have its square root found. 



Find 


V4 


hank roving. 


Find 


V 5 


hank roving. 


Find 


V 6 


hank roving. 


Find 


V7 


bank roving. 


Find 


V8 


hank roving. 



SPINNING ROOM PROBLEMS 

The number of ends that will lie side by side in 
one inch is found by extracting the square root of 
the number of yards of that size yarn in one pound. 

A No. 28 's yarn has 28 X 840 = 23520 yards in 
one pound. How many ends will lie side by side in 
one inch? 

V23520 

98 



A No. 30 's yarn has 30 X 840 = 25200 yards in 
one pound. How many ends will lie side by side in 
one inch? 



V25200 

Find the square root of the following numbers 
of yarns : 

VslT slW, a/22~, Vl2~, 

V36^ V25~, a/32~, V^T 

WEAVE ROOM PROBLEMS (Square Root) 

1. How many ends will lie side by side of a 30 's 



yarn 



V30 X 840 = 



2. How many ends will lie side by side of a 36 's 
warp? 



V36 X 840 = 

3. How many ends will lie side by side of a 22 's 
warp? 



V22 X 840 = 
4. The diameter of a yarn is found by dividing 
•034 by Vn^ 

Find the diameter of a 36 yarn. 

.034 1034 

— — == — — =• .0056 inches. 
V36 6 



99 



Equations 



When we wish to show that several values when 
taken collectively,, either expressed by addition, 
subtraction, multiplication or division, are equiva- 
lent or equal to certain other indicated values this 
is shown in a form known as an equation. Thus, 
8 + 2 + 1 = (equals) 7 -f 2 + 2, or 12 + 2 — 3 
= 6 + 3 + 2, or 3 X 6 X 10 = 5 X 6 X 6 or 

10 X 16 4 X 20 
= are all equations. 



As both sides of an equation are the same, dividing 
or multiplying both sides through by a common 
number does not alter its value. Thus, if we have 

16 X40 

the equation, 16 X 10 = , and divide both 

4 

sides through by 16 the relative values of both sides 
have not been changed, and the remaining equation 

40 

would be 10 = . 

4 

When one of the numbers in an equation is un- 
known it is usually indicated by the letter, X. 
Thus, if we have 10 X X = 15 X 10 we can find 
the value of X by dividing both sides through by 10. 
Then this would give us the equation X = 15. 



100 



EXAMPLES 

The surface speed of two pulleys which are driven 
by a belt must be equal. This surface speed, or 
speed of a point on the outer rim of a pulley is 
found by multiplying the diameter, or distance 
across, by the number, 3.1416, which gives the cir- 
cumference,, or distance round. This in turn mul- 
tiplied by the revolutions per minute of the pulley 
gives the surface speed. 

Example : A 24 inch pulley is running 200 revo- 
lutions per minute and is driving, by a belt, another 
pulley which is 10 inches in diameter. Find the 
speed of the 10 inch pulley. As the surface speeds 
of the two pulleys must be equal (unless belt slips), 
and as surface speed of each pulley is found as given ' 
above we have : 

(a) 24 X 3.1416 X 200 = surface speed 

of 24 inch pulley, and 

(b) 10 X 3.1416 X (X unknown) = sur- 

face speed of 10 inch pulley. 

Then as these two surface speeds are equal we have : 
24 X 3.1416 X 200 = 10 X 3.1416 X (X unknown). 
Then as dividing both sides of an equation does not 
alter its value, we can divide both sides of this one 
through by 3.1416 and our remaining equation is 

24 X 200 = 10 X (X unknown speed). 

Thus we see that in all cases of this kind the num- 
ber, 3.1416, will be left out and we have the general 
principle that the diameter (distance across) of a 
pulley, multiplied by its speed always equals the 



101 



diameter of the other pulley multiplied by its speed. 
In the above remaining problem, then if both 
sides of the equation are divided by 10 the value 
is not changed and we have 

24 X 200 

= X (unknown speed of 10" pulley). 

10 

Then the speed of the 10" pulley would be 480. 

In like manner the diameter of either the driver 
or driven pulleys may be found as well as the 
speeds. 

Example : A 12 inch pulley is to be run 160 rev- 
olutions per minute from a shaft which is running 
192; what size pulley must be put on the driving 
shaft? 

12 X 160 == 192 X (X unknown pulley). 

12 X 160 

Then — = X = 10 inch pulley. 

192 

Example : Driving shaft runs 212 and has a 16 
inch pulley driving a 12 inch pulley; what is speed 
of 12 inch pulley? 

212 X 16 = 12 X (X unknown speed) ; 

212 X 16 
Then X = . 



12 

When one gear is driving another the relative 
speeds of the two are determined from the number 
of teeth in each gear. 

Thus if a gear having 50 teeth is driving an- 
other of 50 teeth the driver in turning one time 

102 



will take up the 50 teeth in the driven gear, thus 
causing- it to make one turn at the same time. If 
the driver makes 100 turns in one minute, then 
the driven will also make 100 turns in the same 
time. If, however, we have 50 teeth in the driver 
and only 25 in the driven, the driver making one 
turn must take up 50 teeth in the driven gear. But 
as this gear only has 25 teeth we readily see that 
it will have to turn two times while the driver of 
50 teeth is turning one time. Then the relative 
speeds of two gears is found by dividing the num- 
ber of teeth in larger by the number of teeth in the 
smaller. Thus if we have a 160 teeth gear driving 
a 40 teeth gear the relative speeds of the two is 
found by dividing 40 into 160, which goes 4. This 
means that while the 160 teeth gear is making one 
turn the 40 teeth gear will make 4 turns. Again we 
see that if this 160 teeth gear makes 10 turns in a 
minute the total number of teeth passing by a cer- 
tain point in one minute will be 10 X 160 = 1600. 
If then this gear meshes into the 40 teeth gear it 
will cause this same number of teeth from this gear 
to pass the same point. Then as this 40 teeth gear 
must deliver 1600 teeth to this point in the same 
time it must turn 40 divided into 1600 = 40 turns 
in one minute. From the above then we can see 
that when two gears are meshing, the number of 
teeth in one gear multiplied by its speed must equal 
to the number of teeth in the other gear multiplied 
by its speed. Suppose we have a gear of 40 teeth 
running 150 revolutions per minute meshing into 
another gear of 30 teeth and wish to know how 
fast the 30 teeth gear is turning. Then we have the 



103 



equation, 40 X 150 = 30 X (X unknown speed). 

40 X 150 

Therefore, by solving we have = X, the 

30 

speed of 30 teeth gear = 200 speed required. 

In like manner we can find the speed of either 
gear thus meshing or the size of either gear, to pro- 
duce a certain speed. 

Example: A gear has 40 teeth and is running 
116 turns per minute; what size gear must be used 
with it to produce 145 turns per minute ? 

Equation, 40 X 116 = 145 X (X unknown gear) ; 

40 X H6 

then, = X = 32 gear required. 

145 

In solving equations, long processes are avoided 
by substituting equal values for certain numbers. 

Example : If the back roll in a machine takes in 
20 inches in a minute and front roll delivers 200 
inches in a minute, what is the draft and what the 
revolutions per minute of front 1" roll ? 

As front roll delivers 200 inches while back roll is 
taking in 20 inches, the front roll is delivering 200 
-f- 20 = 10 times as much as the back roll. This is 
draft, or the draft is 10. 

Surface speed of front roll 

Then, draft = . 

Surface speed of back roll 

Now as draft in this case is 

200 200 Surface speed of front roll 
we have = : . 



20 20 Surface speed of back roll 

104 



As surface speed of back roll is % that of the front 
roll, we have, 

200 Surface speed of front roll 



20 /4oX (Surface speed of front roll) 

Now as surface speed of a roll = circumference X 
revolutions or turns we have, 

200 circumference X revolutions 



20 Yiq X (surface speed of front roll) 

And as the surface speed of' front roll is 200 inches 
per minute, we have 

200 circumference X revolutions 



20 y 10 X (200) 

Also if circumference = diameter X 3.1416, and as 

200 
the diameter of front roll is 1 inch we have, = 



20 
lX3.1416Xrevolutions lX3.1416Xrevolutions 



Mo X (200) 20 

Because Y 10 of 200 or y 10 X 200 = 20, 

200 IX 3.1416 X revolutions 
then, = . 



20 20 

If both sides of this equation is multiplied by 20 the 
value is not changed and we have, 

20 X 200 20 X 1 X 3.1416 X revolutions 
20 20 

105 



This will do away with the number 20 on both sides 
because the 20 below the line on left goes into the 
20 above the line 1 time, and 1 time 200 1 = 200. 
Also the 20 below line on right hand side of the 
equation will go 1 time into the 20 above the line. 
Then we have, 

200 = 1 X 3.1416 X revolutions. 

Again dividing both sides by 1 X 3.1416 will not 
change the value and we have,. 

200 1 X 3.1416 X revolutions 



1 X 3.1416 3.1416 

And as the 3.1416 above and below the line will can- 
cel each other we have, 

200 

= revolutions, or 



1 X 3.1416 



200 

revolutions = = 63.6 revolutions. 

1 X 3.1416 

From this as a general rule we have, R.P.M., (revo- 

surface speed per minute 



lutions per minute) = 

dia. X 3.1416 

And R.P.M. X diameter X 3.1416 = surface speed. 

FORMULAS 

A formula is an equation showing in a general 
way how a problem is to be solved. Thus from the 
preceding we have the following formulas : 



106 



To find the speed of pulleys when diameters are 
given or visa versa. 

Formula: SpeedXdiameter = speed X diameter. 
( one pulley ) ( other pulley ) 

To find the speed of gears when the number of 
teeth are given or visa versa. 

Formula: Speed X No, teeth = speed X No. teeth, 
(one gear) (other gear) 

CARD ROOM PROBLEMS 

1. If the driving pulley of a card is 20 inches 
in diameter and it is to run 165 revolutions per min- 
ute, what size pulley must be on the line shaft which 
runs 200 revolutions per minute? 

Formula: SpeedXdiameter = speed X diameter: 
Then 165 X 20 = 200 X diameter (pulley on shaft). 
Then solving the equation we have by dividing both 
sides by 200 : 

165 X 20 200 

= X diameter. 

200 200 

And as the 200 above and below the line on right 
will cancel we have : 

165 X 20 

= dia. of pulley on shaft=16.5=16^ in. 

200 

2. If the doffer is running 10 revolutions per 
minute and doffer gear has 192 teeth, how many 
revolutions per minute is the doffer change gear of 
20 teeth making ? 

Formula: Speed X number teeth = speed X 



107 



number teeth; then substituting we have, 10 X 192 
= speed X 20. And, speed = 

10 X 192 

= 96 Rev. per minute of change gear. 

20 

SPINNING ROOM PROBLEMS 

1. If the cylinder on spinning frame is to run 
800 revolutions per minute and has a 12 inch pul- 
ley, what size pulley must be on line shaft which 
has a speed of 300 revolutions per minute? 

Formula : Speed X diameter = speed X di- 
ameter; then substituting we have, 800 X 12 = 
300 X diameter ; then diameter = 

800 X 12 

= 32 inch pulley on shaft. 

300 

2. If the cylinder is turning 800 revolutions per 
minute and has a 26 teeth gear, how fast is the stud 
gear of 132 teeth turning? 

Formula : Speed X number teeth = speed X 
number teeth; then substituting we have, 800 X 26 
= speed X 132, then speed = 

800 X 26 

= 157.5 Rev. per minute of stud gear. 

132 

WEAVE ROOM PROBLEMS 

1. If a loom has a 12 inch pulley and is to run 
160 picks per minute, what size pulley must be on 
line shaft which is running 200 revolutions per 
minute ? 



108 



Formula: Speed X diameter = speed X diame- 
ter ; then substituting we have,, 160 X 12 = 200 X 
diameter, and diameter = 

160 X 12 

. = 9.6 inches Dia. of pulley on shaft. 

200 

2. If an auxiliary harness cam shaft is to be used 
mi a loom to make 1 turn for 6 picks and it has a 60 
teeth gear on it, what size gear must be on the cam 
shaft of loom to drive it? 

As the cam shaft of loom makes 1 turn for every 
2 picks of loom this auxiliary shaft will make 1 
turn for every 3 turns of cam shaft in order to equal 
6 turns of the crank shaft. Then we have the for- 
mula : speed X number teeth = speed X number 
teeth and by substituting,, 1 X 60 = 3 X number 
teeth on cam shaft. Then number teeth = 

1X60 

= 20 teeth gear on cam shaft. 



109 



Cancellation 



Cancellation is a short form of division in which 
both the number divided and the number by which 
we are dividing are reduced by dividing both by a 
common number. Thus, if we have 36 to be divided 

36 
by 6 and indicate it thus : — , we can divide both the 

6 
top number and the bottom number through by 2. 
This will give 18 on top and 3 below. This is done 
in practice as follows : 

18' 

^0 18. 

3 
3 

This reduces the top and bottom number to lower 
terms in which it is simple to divide the one into 
the other. 

Example : Use cancellation and divide the 
product of 20 X 16 X 30 X 4 by the product of 
10 X 8 X 24. This is shown thus : 

2o >< iy> x -0 x 4. 

-=20 

;P X ? X 24 

2 6 

By selecting any number either above or beloAv 
the line which we see is easily divided the one into 
the other we thus divide, striking out, or cancell- 

110 



ing, these two numbers and placing the result of this 
division above or below the number thus divided. 
In the above example we readily see that 4 will go 
into 24 evenly, giving the result, 6. Then we place 
6 below 24 and draw a line through both 4 and 24. 
Again we see that this 6 will go 5 times into 30 
above the line and we place 5 above 30,, striking out 
both 6 and 30. Again this 5 will go 2 times into 10 
and we strike out the 5 and 10 and place 2 below 
10. Also we see that this 2 will go 8 times into 16 
and we place 8 above 16 and strike out 2 and 16. 
Again this 8 will go one time into the 8 below the 
line and these two numbers are cancelled, leaving 
as a result of this division 20 above the line and 1 
below. As 1 goes into 20, twenty times the result 
of our division is 20. 

CARD ROOM PROBLEMS 

1. The doubling on a drawing frame is 6 ends of 
a 50 grain sliver. The Aveight of one yard going in 
back must be 6 X 50. If the draught is 5 what is 
the weight of sliver produced in front? 



10 

ffl X 6 



= 10 X 6 = 60 grains. 



2. The doubling on drawing frame is 6 ends of a 
50 grain sliver. The weight of one yard going in 
must be 6 X 50. If the weight per yard in front is 
desired to be 60 grains what draft must be used? 



5 

xm 



draught = 5 draught. 



00 
) ' 

111 



3. (a) If jack shaft on the slubber makes 290 
revolutions per minute and jack shaft gear has 52 
teeth driving to spindles while gear on spindle shaft 
has 58 teeth, what is the speed of spindle shaft? 

As seen under equations, 290 X 52 = 58 X (X, 
unknown speed). Then, 

290 X 52 

= X, speed of spindle shaft. 

58 

(b) If spindle shaft has gear of 50 teeth driving 
spindle gear of 26 teeth, what is speed of spindle? 

As above, (a), speed of spindle shaft X its gear 
= speed of spindle (X, unknown) X spindle gear. 

Then as speed of spindle shaft is represented by 

290 X 52 

the unsolved part of equation, , if we place 

58 
this number in to represent this speed we have 
from (a) : 

290 X 52 

, (speed of spindle shaft), X 50, 

58 

(spindle shaft gear), = speed of spindle,. (X, 
unknown), X 26, (spindle gear). This gives 

290 X 52 X 50 
= 26 X X, unknown spindle speed. 

58 

Then to solve for-X as in equations we have (di- 
viding both sides through by 26) 

290 X 52 X 50 

= spindle speed. 

58X26 

Then to solve by cancellation we have : 



112 



2 
10 M 25 
200 X&2X £0 

58 X 2(> 10 X 2 X 25 = 500 revolutions per 

20 18 minute spindle speed. 

SPINNING ROOM PROBLEMS 

1. How many pounds of No. 20 's warp yarn per 
spindle per 10 hours will be produced by a frame 
with 1 inch front roll running 100 E. P. M. ? 

5 
5 J0 

1X3.1416XWX00XJ0 5X5X3.1416 

= = .28 pounds. 

36X840X20 36X7 

1 

2. If 3.1416 is the circumference of front roll, 
front roll gear 100 teeth, stud gear 132 teeth, cylin- 
der gear 26 teeth, change gear 20, and ratio of whirl 
to cylinder 7.8, what is the twist per inch? 

66 
m X WX7.8 66X7.8 

= = 63.02 turns per inch. 

3.1416X20X2$ 2.6X3.1416 

ft 

2.6 

WEAVE ROOM PROBLEMS 

1. A loom runs 160 picks per minute, on goods 
with 40 picks per inch ; how many yards will it make 
in 10 hours ? 

i 

jmx 20X10 20X10 
= = 66.66 yards. 

MxW s 

3 

113 



2. A cloth has 72 ends per inch of 30 's warp and 
is 36 inches wide. What is the weight of warp in 
840 yards allowing 6% take up? 

1.2 
72X£0XW 72X1.2 

= 91.9 pounds. 



.94, (1— .O6)XWX00 .94 



114 



Percentage 



As most all items in cotton manufacturing, such 
as production, waste, machines stopped and run, and 
dividends made, are expressed as being a certain 
"per cent," it is necessary for us to first clearly 
understand what "per cent" or "percentage" 
means. The word "per cent" comes from the Latin 
words "per" meaning "by" or "by the" and 
"centum" meaning "one hundred." Thus, the 
word "per cent" means "by the hundred," and 
when Ave say "ten per cent" we mean "ten by the 
hundred" or "ten parts of a hundred." Per cent 
then means the relation of a certain number to the 
number, 100. Thus if we say "ninety-eight per 
cent, ' ' we mean that in every one hundred parts of 
a certain number we are considering 98 parts. If Ave 
run through the mill 100 pounds of cotton and pro- 
duce 88 pounds of cloth Ave sell 88 per cent of the 
cotton bought or put in. If we run 500 pounds of 
cotton and get 88 pounds of cloth from every 100 
pounds, then we will produce 5 X 88 = 440 pounds 
of cloth. Thus we see that multiplying the number 
of pounds made from each 100 by the number of 
hundreds put in will produce the number of pounds 
made. 

As per cent means a certain part of 100, when Ave 
say "5 per cent" it can be expressed thus: 5 / 100 
of 100 or % 00 X 100. Then % 00 X 100 = ™% 00 
= 5. Percentage being thus always expressed 
in hundredths it can be shoAvn as a decimal 



115 



and is almost always so shown. Thus 4 per cent = 
/loo = -04 ; 6 per cent = % o = -06 ; 10 per cent = 
io/ 100 = .10; 15 per cent = 15 / 100 = .15; 25 per 
cent = 2 %oo = -25, etc. 

The sign of "percentage" is made thus, % ; and 
when used after a number it is read "per cent." 
Thus 6% = six per cent = .06 ; 8% = eight per 
cent = .08; 10% = ten per cent = .10; 50% = 
fifty per cent = .50; 100% = one hundred per 
cent = 1.00; 116% = one hundred and sixteen per 
cent = 1.16; 13%% = thirteen and one-half 
per cent = .135; 8%% = 8.75% = eight and three- 
fourths per cent = .0875; 2%% = 2.25% = two 
and one-fourth per cent = .0225; %% = .5% = 
one-half per cent = .005 ; %% = .25% = one-fourth 
per cent = .0025; 2%% = 2.5% = two and one- 
half per cent = .025. 

Thus we see that to reduce a number expressed 
as % to its decimal equivalent we divide by 100. 

Examples: Reduce .253% to decimal equivalent. 

100. ) .25300 ( .00253 decimal equivalent. 
200 

530 
500 

300 
300 

Reduce .052% to decimal equivalent. 
.052 -=- 100 = .00052. 

Reduce .125% to decimal equivalent. 
.125 -=- 100 = .00125. . 



116 



Reduce 2%% to decimal equivalent. 
2%% = 2.375%; 
then 2.375 -=- 100 = .02375. 

Prom these we see that to reduce a number ex- 
pressed in % to its decimal equivalent all that is 
necessary is to move the decimal point two figures 
to the left. 

PROCESSES 

(a) To find the "per cent" (%) of a given num- 
ber, multiply the number by the number of per 
cent expressed in decimals. 

Examples : 1. How much is 20% of 75? 
20% = .20; 

then 75 X .20 = 15.00 (fifteen) ; 
then 15 is 20% of 75. 

2. How much is 22%% of 82.5? 

22y 2 % = 22.5% = .225; 
then 82.5 X -225 = 18.5625 ; 
then 18.5625 is 22%% of 82.5. 

3. Find 8%% of 1012. 

83/ 4 % = 8.75% = .0875; 
then 1012 X -0875 = 88.55 ; 
then 88.55 is 8%% of 1012. 

4. If a mill makes 12% waste and uses 120,000 
pounds of cotton in a month, how many pounds of 
waste does it make ? 

12% = .12 
and 120,000 X -12 = 14,400 pounds waste. 

(b) To find what "per cent" one number, A, is 
of another, B, multiply the first number, A, by 100 
and divide by the second number, B. 

117 



Examples: What % is 2 of 100? 
2 X 100 = 200, 

and 200 -r- 100 = 2% = .02. 

5 is what % of 75 ? 
5 x 100 = 500, 
and 500 -=- 75 = 6.666% = .06666: 

What % of 116 is 52? 
52 X 100 = 5200, 
and 5200 -+- 116 = 44.82% = .4482. 

If a mill opens 252,147 pounds of cotton and 
25,622 pounds of waste are made, what is the % of 
waste ? 

25622 X 100 

= 10.16% = .1016. 

252147 

(c) When a number is given which is a certain 
% of another number, to find this number, multiply 
the given number by 100 and divide by the % (not 
in decimals). 

Examples : If 50 is 20% of a number, what is 
the number? 

50 X 100 = 5000, 

and 5000 -f- 20 = 250 the number. 

This can be proved as follows : 

If 250 is the number then what 

number is 20% of it? 
250 X -20 (20%) = 50, 
or if 250 is the number, 50 is 

what % of it? 
50 X 100 = 5000, 
and 5000 ~ 250 = 20%. 



118 



2. What is the number if 116 is 5y 2 % of it? 

116 X 100 = 11600, 

and 11600 -^- 5.5 (5y 2 %) = 2109.09. 

3. 30% of a certain number is 416. What is the 
number? 

416 X 100 = 41600, 

and 41600 -4- 30 = 1386.6. 

4. A mill produces in one month 187,.516 pounds 
of cloth, which is 96.7% of what it could produce. 
How much could it produce? 

187516 X 100 

= 193915.1 pounds. 

96.7 

(d) When a number, A, is given which is the 
result after allowing a certain % loss on another 
number, B; to find, the number, B. 

Subtract the % (in decimals) from 1.00, and 
divide the result into the number, A. 

Example: If 81 is the result after allowing 10% 
loss, what is the number? 
1.00 — .10 = .90. 
Then 81 -f- .90 = 90. 

Then 81 -^ .90 = 90. 

Proof : If 90 is the number then 10% of 90 = 
90 X -10 = 9, 
and 90 — 9 = 81. 

Example : If 10% waste is made how much cotton 
was used in making 18000 pounds of cloth? 
1.00 — .10 = .90. 

18000 -f- .90= 20000 pounds. 



119 



(e) When a number, A, is given which is the 
result after allowing a certain % gain on another 
number, B ; to find the number, B. 

Add the % (in decimals) to 1.00 and divide the 
result into the number, A. 

Eample : If 88 is the result of a number's gaining 
10%, what is the number? 

1.00 + .10. = 1.10. 
88 ~ 1.10 = 80. 
Proof: If 80 is the number then 10% of 80 = 
80 X .10 = 8, 
and 80 + 8 = 88. 

Example : If a loom beam has 109 pounds of sized 
warp on it, what was the weight of unsized yarn 
if 10% is put on? 

1.00 + .10 = 1.10 

109 -f- 1.10 = 99.09 pounds. 

CARD ROOM PROBLEMS 

1. If a picker makes 4% waste, how many pounds 
of Avaste would you expect from running 22,412 
pounds of cotton? 

22412 X -04 = 896.48 pounds of waste. 
(See a.) 

2. In above example, if only 710 pounds of waste 
is collected, what is the % of "invisible" waste? 
896.48 — 710 = 186.48 pounds of invisible waste; 

186.48 X 100 

then (see b) = .83% invisible waste. 

22412 

The "actual" waste being 710 pounds the % is 

710 X 100 

(see b) = 3.16% actual waste. 

22412 

120 



What is "invisible" waste? 

3. A man receives $.85 which is a 3% premium on 
what he makes in a week. What is his actual wage ? 

.85 X 100 

(see c) = $28.33 actual wage. 

3 

4. A card room has 2516 speeder spindles and 412 
are stopped for a week. What is the % of spindles 
run? 

2516 — 412 = 2104 spindles run ; 

2104 X 100 

= 83.6% spindles run. 



2516 
SPINNING ROOM PROBLEMS 

1. A mill has 42,168 spindles and 5% are stopped 
for a week; how many spindles stopped? 

42168 X -05 (see a) = 2108.40 spindles stopped. 

How many spindles run? 

42168 — 2108 = 40060 spindles run. 

What is the % of spindles run? 

40060 X 100 

(see b) = 95% spindles run. 

42168 

2. If a frame is figured to put in 25 turns per inch 
in a certain yarn and the contraction in length due 
to twisting is 3%, what is the actual turns per inch 
put in? 

25 X .03 = .75 turns extra on account of twist ; 
then 25 -f- .75 = 25.75 actual turns per inch. 

121 



3. The yarn from a frame numbers 20 's. What 
is the % of contraction if a 3 hank roving is used 
with a 6.87 draft? The number of yarn ought to 
be 6.87 X3 = 20.61. As,, however, it actually num- 
bers 20 's, which is heavier than 20.61 's, there must 
be some contraction in twisting. 

20 X 100 

Then = 97%, 

20.61 

which means that the resulting number is 97% as 
long as the untwisted strand would be. 

Then the % of shrinkage is found by subtracting 
.97 from 1 which gives 3% contraction. 

4. If a frame is calculated to make 52 pounds per 
day after allowing 2% off for doffing, oiling, etc., 
what is the figured production? 

If 52 is the result after deducting 2%, then 52 
pounds is 98% of the whole. 

52 X 100 
Then = 53.06 figured production. 

98 

WEAVE ROOM PROBLEMS 

1. A warp is set 39.5 inches in reed and makes 
37.5 inch cloth; what is the % of shrinkage? 

39.5 - — 37.5 = 2 inches shrinkage ; 

2 X 100 

then = 5.06% shrinkage. 

39.5 

2. The cut marks on a warp are 68 yards apart. 
Cloth made from this produces 64 yard cuts. What 
is the % "take up?" 

122 



68 — 64 = 4 yards take-up ; 

4 X 100 

then = 5.88% take-up. 

68 

3. If cloth cuts are to be 60 yards and 6% is al- 
lowed for the take-up, how far apart must the cut 
marks be placed? 

If 60 yards is 6% less than the warp length then 
it would be 94% of the warp length. 

60 X 100 

Then = 63.83 yards between cut marks. 

94 

60 

or = 63.83. 

1.00 — .06 

4. A cloth is to be made 36.5 inches wide. If 7% 
shrinkage is allowed, how wide must the warp be 
in reed? 

If 36.5 inches is 7% less than the width in reed, 
it is 93% of this width. Then we have, 

36.5 X 100 

= 39.24" Avidth in reed. 

93 

36.5 

or = 39.24" 

1.00 — .07 

5. Cut of cloth weighs 12.4 pounds and has 7.46 
pounds of warp in it. What is the % of warp and 
filling in the cloth? 

6. If a mill produces 146,218 pounds of cloth 
which has 68% of warp, how many pounds of warp 
and filling are in this amount of cloth? 

123 



Weights and Measures 

There are three principal standard measures used 
in cotton manufacturing. These are weights, 
lengths and time. 

WEIGHTS 

The system of weights used is that of the stand- 
ard pound in which there are 16 ounces. 

1 pound (lb) =16 ounces (ozs.). 
For weighing small amounts of yarn, cloth or 
cotton, the weights are usually obtained in grains, 
and for fractional parts of grains the decimal sys- 
tem is used. Thus, we may have 136.25 grains of 
yarn or cloth, 18.47 grains of yarn, etc. In order to 
use smaller numbers it sometimes becomes neces 
sary to change from one system of weights to an- 
other ; i.e., from grains to pounds or ounces, or from 
pounds to ounces or ounces to grains. To do this 
we must bear in mind the following relation : 

1 pound (lb) =16 ounces (ozs.) =7000 grains (grs.) 

If there are 7000 grains in 1 pound then in 2 
pounds there would be 2 X 7000 = 14000 grains. 

Then to reduce pounds to grains we have, 
No. pounds X 7000 = grains. 

Examples : How many grains in y 2 pound ? 
y 2 X 7000 = 7000 / 2 = 3500 grains. 

How many grains in % pounds? 

% X 7000 = 49000 / 8 = 6125 grains. 



124 



How many grains in % pounds? 

% X 7000 = 2100 % = 5250 grains. 

How many grains in 1.37 pounds? 
1.37 X 7000 = 9590 grains. 

If 7000 grains = 1 pound = 16 ounces, then the 
number of grains in 1 ounce can be found by divid- 
ing 7000 by 16. Thus, 7000 -j- 16 = 437.5 grains 
in one ounce. Then 

No. ounces X 437.5 = grains. 

Examples : How many grains in 2 ounces ? 
437.5 X 2 = 875.0 grains. 

How many grains in 12 ounces ? 

437.5 X 12 = 5250 grains. 

How r many grains in 3.45 ounces? 

3.45 X 437.5 = 1509.375 grains. 

Again, if 1 pound = 7000 grains, then in 14,000 
grains there would be 

14000 

■ — = 2 pounds. 

7000 

Then to bring grains to pounds we have, 

No. of grains 

= No. pounds. 



7000 

Examples : How many pounds in 15,640 grains ? 
15640 -f- 7000 = 2.234+ pounds. 

How many pounds in 4860 grains? 

4860 U- 7000 == .6942 pounds. 



125 



Also if 1 ounce = 437.5 grains, then in 875 grains 
there would be 

875 
= 2 ounces. 



437.5 
Then to reduce grains to ounces we have 

No. of grains 

= No. ounces. 



437.5 

Examples : How many ounces in 1575 grains ? 
1575 -=- 437.5 = 3.6 ounces. 

Hoav many ounces in 4562 grains? 
4562 -f- 437.5 = 

LENGTHS 

In measuring lengths the standard of 1 yard is 
taken in which there are 36 inches. 

When Ave have yards to change to inches all that 
is necessary is to multiply the number of yards 
by 36. 

No. of yards X 36 = No. inches. 

Examples : Hoav many inches in 2 yards ? 
2 x 36 = 72 inches. 

How many inches in 3.75 yards? 

3.75 X 36 = 135 inches. 

When Ave have inches to change to yards we divide 
the number of inches by 36. 

No. of inches 

= No. of yards. 

36 

126 



Examples: How many yards in 416.84 inches? 
416.84 -=- 36 = 11.578 yards. 

How many yards in 5624.92 inches? 

5624.92 -r- 36 = 156.24 yards. 

CIRCULAR MEASURE 

, The distance around a circle, such as a roll or 
cylinder, is found by multiplying the diameter 
(distance through) by 3.1416. 

Examples : A roll measures 1% inches in diame- 
ter; how many inches around (circumference)? 

1% inches = 1.25 inches ; 
then 1.25 X 3.1416 = 3.927 inches circumference. 

If a cylinder measures 27% inches in diameter, 
what is its circumference? 

27% = 27.25 ; 
then 27.25 X 3.1416 = 85.6086 inches circumference. 

TIME 

The measure of time is based on the standard of 
hours in which there are 60 minutes. 

Examples : How many minutes in 60 hours ? 
60 X 60 = 3600 minutes. 

How many hours in 672 minutes ? 

672 -=- 60 = 11.2 = liy 5 hours. 

CARD ROOM PROBLEMS 

1. If a roll delivers 637.42 inches in 1 minute, how 
many yards will it deliver in 10 hours? 



127 



637.42 -=- 36 = 17.7 yards delivered in 1 minute. 

1 hour = 60 minutes; 

then 10 hours = 10 X 60 = 600 minutes. 

17.7 X 600 = 10620.0 yards delivered in 10 hours. 

2. If a roll delivers 436.42 inches of sliver weigh- 
ing 56 grains per yard in 1 minute, how many 
pounds A\ r ill it deliver in 10 hours? 

436.42 -=- 36 = 12.122 yards delivered in 1 minute ; 

then 12.122 X 56 = 678.832 grains delivered 

in 1 minute. 

10 X 60 = 600 minutes in 10 hours. 

Then 678:832 X 600 = 403699.200 grains 

delivered in 10 hours. 

And, 403699.2 -r- 7000 = 57.67 pounds 

delivered in 10 hours. 

3. If a roll 1% inches in diameter is running 200 
revolutions per minute, how many pounds of a 60 
grain sliver will it deliver in 1 hour? 

1% inches = 1.375 inches ; 

1.375 X 3.1416 = 4.3197" circumference. 

If it turns 200 times in 1 minute it will deliver 

4.3197 X 200 = 863.9400 inches per minute; 

863.94 -T- 36 = 23.97 yards delivered per minute. 

1 hour = 60 minutes, 

then 23.97 X 60 = 1438.20 yards delivered in 1 hour, 

then 1438.2 X 60 = 86292 grains delivered in 1 hour, 

and 86292 -f- 7000 = 12.327 pounds 

delivered in 1 hour. 

4. If a lap weighs 12 ounces, how many grains 
would it weigh? 



128 



437.5 grains in 1 ounce; 
then 437.5 X 12 = 5250 grains. 

5. If 100 yards of sliver weigh 52 grains per 
yard, what would it be in ounces? 

100 X 52 = 5200 grains in 100 yards ; 
then 5200 -f- 437.5 = 11.88 ounces. 

6. If a roving can containing 56 a grain sliver 
weighs 12 pounds, how many yards of sliver in it? 

12 X 7000 = 84000 grains weight of sliver. 
Then 84000 -=- 56 = 1500 yards. 

SPINNING ROOM PROBLEMS 

1. If a roll is 3.14 inches in circumference and is. 
turning 100 r.p.m., how many yards will it deliver 
in 1 hour? 

3.14 X 100 = 314 inches delivered in 1 minute ; 
then 314 X 60 = 18840 inches delivered in 1 hour,. 

18840 

and = 523.3 yards delivered in 1 hour. 

36 

2. If 120 yards of yarn weigh 60 grains, how many 
pounds would there be in 22,860 yards? 

120 : 60 : : 22860 : X 

22860 

X = 60 X = H430 grains 

120 
weight of 22860 yards. 

Then, 11430 -f- 7000 = 1.63 pounds. 



129 



3. If bobbins are 1% = 1.25 inches in diameter, 
how many would lie side by side on 6 feet on the 
floo.- ? 

6 feet = 6 X 12 = 72 inches, 
and 72 -=- 1.25 = 57.6 or almost 58. 

4. If 1 bobbin contains 967.5 grains of yarn, how 
many pounds would 330 bobbins weigh? 

967.5 X 330 = 319275 grains in 330 bobbins ; 
then 319275 -=- 7000 = 45.6 pounds. 

WEAVE ROOM PROBLEMS 

1. If 6 inches of 34" cloth weigh 142 grains, how 
many yards would there be in 1 pound? 

6 : 142 : : 36 : X 

142 X 36 

X = = 852 grains 

6 

weight of 36" or 1 yard of 34" cloth ; 

then if 1 pound = 7000 grains, 
7000 -H- 852 = 8.21 yards per pound. 

2. If a shuttle travels 460 inches in a second, how 
many feet would it go in 1 minute ? 

60 seconds = 1 minute ; 
then 460 X 60 = 27600 inches in 1 minute, 
and 27600 -=- 12 = 2600 feet in 1 minute. 

3. A certain loom makes 2.5 inches of cloth per 
minute ; how many yards would it make in 60 hours? 



130 



2.5 X 60 = 150.0 inches made in 1 hour ; 

150 X 60 = 9000 inches made in 60 hours, 

and 9000 -=- 36 = 250 yards in 60 hours. 

4. If a cloth weighs 3.72 ounces per yard, how 
many pounds ought 60 yards to weigh? 

3.72 X 60 = 223.20 ounces, 
and 223.20 -=- 16 = 13.95 pounds. 



131 



Mechanical Calculations 



AREA 



A square inch is a surface area which is, or is 
equal to, a surface which measures one inch on all 
four sides of a square. 





To find the number of square inches in figures 
like those shown in Fig. 2 multiply together the 
length of two sides. If the figure is like (b) a short 
side, A, and a long side, B, is multiplied together. 

Example : If a square measures 14 inches on all 
sides, how many square inches does it contain? 
14 X 14 = 196 square inches. 

A square yard is a surface area equal to the 
product of multiplying 36 X 36 = 1296 square 
inches. 

To change square yards to square inches multiply 
by 1296. 

Example : How many square inches in 2.5 square 
yards? 

1296 X 2.5 = 3240 square inches. 

To change square inches to square yards divide 
by 1296. 



132 



Example : How many square yards in 5642 
square inches? 

5642 -f- 1296 = 4.35 square yards. 

A square foot is obtained by multiplying 12 X 12 
= 144 square inches. 

To change square feet to square inches multiply 
by 144. 

To change square inches to square feet divide by 
144. 

To find the area (square inches) in a figure like 
(b) Fig. 2, multiply a short side by a long side, or, 
A X B. 

Example : How many square inches in a piece of 
cloth 36 inches long and 28 inches wide? 

36 X 28 = 1008 square inches. 






© © 

FIG. 3. v -^ 

To find the area of a figure like (a) Fig. 3. 

XXY 
= area 



(square inches if X and Y are in inches). 
To find the circumference (distance round) from 
A back to A again in (b) Fig. 3, multiply the di- 
ameter, X, by 3.1416. 

Example: What is the circumference of a roll 
4 inches in diameter? 

3.1416 X 4 = 12.5664 inches. 



133 



To find the area of a disk or the square inches 
contained in a circle like (c) Fig. 3. 

Square the diameter, X, (multiply by itself) and 
then multiply by .7854 (% of 3.1416). (Because 

(X) 2 X 2 

area = x 3.1416,. X 3.1416 = X 2 X-7854). 

(2) 2 4 

Example : How many square inches in a spool 
head 5 inches across? 

5 X 5 = 25 (5 squared or 5 2 ), 
then 25 X .7854 = 19.635 square inches. 

VOLUME 

When we have a square block which measures 1 
inch on all edges it is called a cubic inch. If 
the block is 2 inches on all edges it will contain 
2X2x2 = 8 cubic inches. This is true because 
the area of a square 2 inches on all sides will be 
2X2 = 4 square inches. If we build this area up 
to 1 inch in height, we have a volume of 4 cubic 
inches, and if we build again so as to make the 
whole 2 inches deep we double the volume and have 
a block which measures 2 by 2, by 2 inches and a 
volume of 8 cubic inches. 

Square Tanks 

Then to find the volume of a square shaped vessel 
we multiply the two sides together, then multiply 
by the height. 



134 





Example : Find the volume in cubic inches of 
tank like (a) Fig. 4, in which the two sides, 
X and X, measure 22" each, and the height, or 
depth, Y, is 52 inches. 

22 X 22 = 484, and 484 X 52 = 25,168 cubic inches. 

A standard gallon contains 231 cubic inches. How 
many gallons would the above tank hold? 
25168 -i- 231 = 108.9 gallons. 

One cubic foot = 12 X 12 X 12 = 1728 cubic 
inches; how many cubic feet in the above tank? 
25168 -r- 1728 = 14.56 cubic feet. 

Example : Find the volume in cubic inches of a 
block or tank like (b) Fig. 4, when the two sides 
X and X measure 22 and 46 inches respectively and 
the height or depth, Y, is 10 inches. 
22X46 = 1012, and 1012X10 = 10120 cubic inches. 

How many cubic feet will it contain ? 

10120 -v- 1728 = 5.85 cubic feet. 

How many gallons will it hold? 

10120 -r- 231 = 43.8 gallons. 

If 1 cubic foot = 1728 cubic inches and there are 
231 cubic inches in a gallon then one cubic foot 



135 



= 1728 -4- 231 = 7.48 gallons. One gallon of water 
weighs 8.35 pounds; then one cubic foot of water 
will weigh 

7.48 X 8.35 = 62.5 pounds. 

Example: If a tank like .(b) Fig. 4 is 3 feet by 
8 feet and 3 feet deep, how many cubic feet will it 
contain ? 

3 X 8 = 24 and 24 X 3 = 72 cubic feet. 

How many gallons will it hold? 
72 X 7.48 = 538.56 gallons. 

How many pounds of water will it hold? 
538.56 X 8.35 = 4496.976 pounds. 

Circular Tanks 

When we have a tank or cylinder like (c) Fig. 4, 
the volume is found by first finding the area of the 
top or the bottom, X, and multiplying this by the 
depth or height, Y. 

Example : Find the volume of a circular tank 
which is 54 inches deep and is 24 inches in diameter. 

Area of bottom = (24) 2 X -7854 = 576 square 
inches. 

Then 576 X 54 = 3104 cubic inches. 

How many gallons will this tank hold? 
231 cubic inches = 1 gallon ; 
then 31104 -f- 231 = 134.6 gallons. 

What will be the weight of water? 

134.6 X 8.35 = 1123.91 pounds. 

To find the volume of tanks like (d) Fig. 4, in 
which the top X is larger than the bottom X, and 



136 



viee versa. Find the average diameter by subtract- 
ing the smaller from the larger and adding one-half 
of this difference to the smaller. Then find the 
area of this average diameter and multiply by the 
depth of tank. 

Example : If top of tank is 34 inches in diameter 
and bottom is 28 inches in diameter what is its 
volume if depth is 30 inches ? 

34 _ 28 = 6, and % = 3. Then average 
diameter is 28 -j- 3 = 31 inches. 

Then (31) 2 X -7854 = (31X31) X -7854 = 754.7694 

square inches, and 754.7694 X 30 = 22643.0820 

cubic inches. 

How many gallons will it hold? 

22643.082 -r- 231 = 98 gallons. 

SPEED OR VELOCITY 

Speed = distance X time. 

There are two kinds of speeds as generally con- 
sidered although they are closely related. These 
are rotary speed and surface speed. Rotary speed is 
spoken of as "revolutions per minute" (r. p. m.), 
and surface speed as "feet per second," "feet per 
minute," "inches per minute," or "yards per min- 
ute,, second," etc. Thus we say a roll or pulley has 
210 r. p. m., meaning it turns 210 ! times in 1 minute. 
Also the speed of a belt, say, is 2100 inches per 
minute, meaning that a point on the belt is travel- 
ing 2100 inches in 1 minute. The speed or velocity 
of a creek or river can be found by throwing a 
cork or light article on its surface and noting how 
far it will travel in a certain time. Rotary speed 



137 



can be converted to surface speed, as in the case of 
two pulleys connected by a belt. Also surface speed 
can be converted to rotary speed, as in the case of 
an engine piston driving a crank. 

Rotary Speeds 

When two pulleys are connected by a belt, as this 
changes rotary to surface speed the rotary speeds 
of both pulleys must accommodate themselves to 
the resulting surface speed of the belt. If they do 
not, one or the other must slip. Then the surface 
speed produced by both pulleys, no matter what 
their diameter, must be equal. If the circumference 
of a pulley is 36 inches and if we roll it on the floor 
for one complete turn it will pass over 36 inches on 
the floor. If we roll it this distance in 1 second, 
then it passes over 36 inches in 1 second, which is 
its surface speed. If we suspend this pulley and 
pass around it a belt and turn the pulley 1 turn in 
one second it will take up 36 inches of belt in 1 
second, which is its surface speed. Then to change 
rotary speed to surface speed we multiply the rotary 
speed by the circumference of the roll or pulley. 
(The circumference = 3.1416 X diameter.) 

Example : What is the surface speed of a pulley 
12 inches in diameter and running 160 r. p. m. ? 

12 X 3.1416 X 160 = 6031.872 inches per minute; 

6031.872 -~ 12 = 502.65 feet per minute ; 

502.65 X 60 = feet per second. 

If in the above example this 12 inch pulley is con- 
nected by a belt to a 14 inch pulley we find the 
speed (rotary) of this pulley as follows : 



138 



As the surface speeds of both pulleys must be 
equal; then 
12 X 3.1416 X 160 = 14 X 3.1416 X (X unknown). 

Dividing both sides by 3.1416 we always eliminate 
this number 3.1416 and we have 

12 X 160 = 14 X (speed unknown). 

Always, then, we have this fact : Speed of a pul- 
, ley X its diameter = speed of other pulley X its 
diameter. 

Then from equations we see that speed of 14 inch 

12 X 160 

pulley = = 137.1 r. p. m. 

14 
Using this equation we can find the sizes or speeds 
or any pulley desired. 

Example : A 9 inch pulley is running 200 r. p. m. 
and is required to drive another which is to run 160 
r. p. m. What must be the size of other pulley? 

9 x 200 = (X) X 160 

9 X 200 

then (X) = = 11.25 = 11% inches 

160 
diameter of other pulley. 

Surface Speeds 

Surface speeds can be converted to rotary speed, 
as in the case of an engine piston driving an engine 
shaft by means of a crank. If an engine fly wheel 
is turning 200 r. p. m.„ the piston having a stroke of 
2 feet, the piston will make two strokes for each 
turn of the crank and fly wheel. 

Then the piston will have a surface speed of 
2 X 2 X 200 = 800 feet per minute. 



139 



About 3500 feet per minute is a good average 
speed for main driving belts. Then if we have a 
motor running 1200 r. p. m., what size pulley should 
it have so as not to give a greater surface speed to 
the belt than 3500 feet per minute? 

Surface speed = diameter X 3.1416 X r. p. m. 
Then 3500 = diameter X 3.1416 X 1200. 

Dividing both sides by (3.1416 X 1200) we have 

3500 
= diameter in feet =.928 ft. diameter 



3.1416 X 1200 

and .928 X 12 = 11.1 inches diameter of pulley. 

Example : If a machine has a 16 inch driving 
pulley which is to be run 320 r. p. m., what surface 
speed will the belt have ? 

16 X 3.1416 X 320 = 16084.99 inches per minute, 
and 16084.99 -f- 12 = 1340.4 feet per minute. 
Is this too high? 

POWER 

Power is the energy required to do a certain 
amount of work in a given time. Work is the result 
of moving a certain weight for a certain distance. 
Then power takes into consideration, weight, dis- 
tance and time. If we raise a block weighing 1 
pound a distance of 1 foot the work done is ex- 
pressed in feet and pounds and this is said to be 1 
foot-pound. If we raise the same block a distance 
of 2 feet, the result is 2 foot-pounds ; or if we raise 
a weight of 2 pounds a distance of 1 foot, it is said 
to be 2 foot-pounds. If we raise a weight of 2 



140 



pounds a distance of 2 feet the result is 4 foot- 
pounds. Thus we see that work or foot-pounds 
= weight X distance. 

As speed = distance in a given time, 
power = weight X speed in a given time. 

Example : If it be found to require a pull of 50 
-pounds to start up a machine and the speed of driv- 
ing belt is 1500 feet per minute, how much power 
is required to run it? 

Power = weight X speed in a given time, 

then power = 50 X 1500 = 75000 

foot-pounds per minute. 

The standard for horse power (H. P.) is, 
1 H. P. = 33000 foot-pounds per minute or 

33000 

1 H. P. = = 550 foot-pounds per second. 

60 

Then in the above example if it requires 75,000 
foot-pounds to move the machine with a belt speed 
of 1500 1 feet per minute, the horse power will be 
75000 -=- 33000 = 2.27 horse power. 

From the above the power required to drive any 
machine may be determined. The following is 
given as a simple method of finding the horse power 
required to run a machine : Pass a cord around the 
driving pulley of machine and give it several turns 
so that when the cord is pulled the machine will be 
started. Then attach to the end of this cord a re- 
liable spring balance. A pull on this spring with 
sufficient force to turn the machine will show the 
weight on balance. Then multiply this weight by 



141 



the working speed of belt in feet per minute and 
divide by 33000. 

Example : A machine has a 12 inch pulley which 
is to run 160 r. p. m. The tension as shown on the 
spring is 22 pounds; how much H. P. will be re- 
quired to run it? 

The surface speed of driving belt will be, 
12 X 3.1416 X 160 = 6031.872 inches per minute, 

or 6031.872 -+- 12 = 502.656 feet per minute ; 

502.656 X 22 
then = .335- horse power, or about 

33000 y 3 H. P. 

Steam Power 

The pressure of steam is expressed in pounds per 
square inch. Thus if steam is admitted to an engine 
cylinder with an average pressure of 30 pounds per 
square inch and the piston head has an area of 122 
square inches, the total pounds pressure exerted 
against the piston is 122 X 30 = 3660 pounds. 

An average coal consumption is about 23 pounds 
per hour for every square foot of grate surface. 
Thus if four boilers have a grate surface of 18 
square feet, the coal consumption will be near 
18 X 23 = 414 pounds per hour, and 414 X 10 
= 4140 pounds per day of 10 hours. One pound of 
coal will evaporate about 8% pounds of water. 

Hence in the above where 4140 pounds of coal are 
used in 1 day it will require about 8% X 4140 
= 35190 pounds of water and as water weighs near 
8y 2 pounds per gallon this will be 4140 gallons of 
water consumed in a day. An average water con- 
sumption is 20 pounds for each indicated horse 



142 



power (1 H. P.). Then if the above boiler consumes 
35,190 pounds of water, the H. P. produced should 
be 35190 -r- 20 = 1709.5. 

To find the indicated horse power of a steam en- 
gine the f ollowing formula may be used : 

2RASP 

H. P. = in which 

33000 

H. P. = Indicated horse power. 
E. = Revolutions per minute of shaft or fly wheel. 
A = Area of piston in square inches. 
S = Stroke in feet. 

P = Average steam pressure in pounds per square 
inch. 

It must be remembered that the indicated horse 
power is always more than the actual horse power 
on account of friction in the engine itself. Thus the 
"efficiency" of an engine is the ratio of actual 
horse power to indicated horse power. Thus Ave 
have, 

A. H. P. 



I. H. P. 

in which A. H. P. means actual horse power. For 
instance, if the indicated horse power of an engine 
is 250 and 235 the actual horse power, then 
23% B0 = .94 = 94^, efficiency. 

The average steam pressure being used in an 
engine is found by the use of an instrument known 
as an "indicator." The application of this instru- 
ment to the engine is familiar to most all engineers. 

After the tracing is obtained on an indicator 
card, the pressure is found as shown by following: 



143 



If the area of card is 6 square inches and it is 5 
inches long, while the scale of indicator spring is 
Y 3 q the steam pressure would be 

6 X30 

= 36 pounds per square inch. 

5 

Example : Find the indicated horse power of an 
engine which has an 18 inch cylinder (diameter), 
a stroke of 32 inches and is running 80 r. p. m., if 
the steam pressure is 36 pounds per square inch. 
Allow for a 2" piston. Using the above given for- 
mula we have, 

11=80 revolutions per minute. 

A= (18X18) X -7854=254.469 square inches. 

As this area is less on the piston rod side, we find 
the area of piston rod and take half of this area 
from the piston head area to give the effective area. 
Then area of piston = 

(2 X 2) X -7854 

= 1.5708, 

2 

then 254.469 — 1.5708 = 252.8982 square inches. 
S = 32 inches = 32 -f- 12 = 2.666 feet. 
P = 36 steam pressure. 

Then 

8 12 

2X^X252.89X2.666X^ 
H. P. = = 117.6 H. P. 



550 



144 



Water Power 

The pressure -exerted by a head of water on a 
given area is equal to the weight of a volume of 
water Avhose area is equal to the area considered, 
multiplied by the height of this column from the 
given area. Thus the weight of a volume of water 
whose area is equal to 1 square inch would be the 
weight of 1 cubic inch of water, which is 62.5 -4- 144 
(as 1 cu. ft. = 62.5 lbs. 1 cu. ft. = 144 cu. inches) = 
.434 pounds. Then if the height of this water (head) 
above a certain point is 40 feet the pounds per square 
inch would be .434 X 40 = 17.360 pounds per square 
inch. It makes no difference what the total volume 
of water at the head is, if we multiply the distance 
below a tank or reservoir by .434 the result is always 
(shown by example) pounds per square inch. 

The weight of a cubic foot of water being 62% 
pounds, if we know the volume of water passing a 
certain point in a stream in a given time we can find 
the foot-pounds and hence the horse power devel- 
oped. Example, if a stream passes 2162 cubic feet 
of water in a minute at a 20 foot fall we will have 
2162 X 62.5 X 20 = 2702500 foot-pounds per minute 
of power developed. Then as 33000 foot-pounds per 
minute = 1 H. P. we have, 2702500 -4- 33000 = 81.8 
H. P. developed. As, however, water turbines do 
not give more than 75% mechanical efficiency, then 
81.8 X -75 = 61.35 effective horse power. 

PUMPS 

The number of gallons of water or other liquid 
delivered by a pump is found by obtaining the vol- 
ume of the cylinder in cubic inches and dividing by 



145 



231. This gives the number of gallons for each 
stroke. Then the number of strokes per minute 
multiplied by this will give the gallons capacity of 
the pump in 1 minute. If a pump has a 4 inch cylin- 
der and a 6 inch stroke, and is running 40 strokes 
per minute, how many gallons will be pumped in 1 
minute ? 

4 X 4 = 16, then 16 X -7854 = 12.5664 
square inches area of piston or cylinder. 

12.5664 X 6 (stroke) = 75.3984 
cubic inches per each stroke. 

Then 75.3984 X 40 = 3015.936 cubic inches 
pumped in 1 minute, and 3015.936 -h 231 (eu. in. 
per gal.) = 13.05 gallons pumped in 1 minute. 

To find the power required to run a pump hav- 
ing a certain capacity and lifting the water a cer- 
tain height multiply the number of gallons pumped 
per minute by 8% (lbs. of 1 gallon) and multiply 
this by the height pumped and divide this by 33000. 
This will give the theoretical horse power. To this 
must be added an allowance for pipe friction, loss 
in pumps, gears, belts, etc. 

Example : What horse power will be required to 
run a pump with a capacity of 500 gallons per min- 
ute pumped to a height of 50 feet ? 

500 X 8^4 = 4250 pounds per minute. 

4250 X 50 = 212500 foot-pounds per minute. 

And 212500 -=- 33000 = 6.43 H. P. Theoretical. 

ELECTRIC POWER AND LIGHTING 

The terms used in electricity correspond to those 
used in mechanical engineering and when thus com- 
pared are more easily understood. . 

146 



Ampere 

Means the amount of current in a wire, light or 
motor and is equivalent to weight as commonly 
used in mechanics. 

Volts or Voltage 

Means the rate of flow of an electric current in a 
„wire, light or motor and is equivalent to velocity, 
speed or "feet per minute" in mechanics. 

Ohm 

Is the unit of resistance as set up against a cur- 
rent by its conductor and is equivalent to loss due 
to friction in mechanics. 

In mechanical power the work performed is found 
by multiplying the weight by the speed in a certain 
time. So in electric power the work performed is 
found by multiplying the amperes by the volts. 
This product is called watts, which represents the 
unit of electric power. Then we have, 

Electricity : Amperes X volts = watts. 
Mechanics : Weight X velocity = power. 

One electrical horse power is equal to 746 watts. 
One kilowatt is equal to 1000 watts. So one kilo- 
watt would be equal to 100 %46 = 1-34 horse power. 

And one horse power would equal .746 kilowatt 
or 746 watts. One watt-hour means one watt ap- 
plied for one hour and is equal to % 46 horse power 
applied for one hour. 

One kilowatt-hour means one kilowatt applied for 
one hour and is equal to 1.34 horse power applied 
for one hour. 



147 



One horse power-hour means one horse power 
applied for one hour and is eual to .746 kilowatt- 
hours. 

Examples : Find watts generated by a dynamo 
having a capacity of 60 amperes and 120 volts. 

120 X 60 = 7200 watts, 

and 7200 ^- 1000 = 7.2 kilowatts. 

As 1 K. W. (kilowatt) = 1.34 H. P. this motor 
would generate 7.2 X 1-34 = 9.648 H. P. 

What current is required for 40, 16-c. p. 3.5 watt 
lamps at 110 volts? 

A lamp at 3.5 watts means that it requires 3.5 
watts for each candle power. 

Then for a 16 c. p. 3.5 watt lamp it would require 
16 x 3.5 = 56 watts. 

Then 56 X 40 = 2440 total watts for all lamps 
= 3.27 H. P. 

2440 watts 

And = 22.18 amperes. 

110 volts 

How many horse power would be required to run 
100, 40 watt lamps? Sometimes the candle power 
of lamps is not given, the term 30, 40 or 50 watts, 
etc., being given. Prom this we can find the candle 
power if we know the usual worth required for each. 
Thus a Mazda lamp requires 105 to 115 volts and 
will produce one candle power for about every .95 
to 1.03 watts. So to solve the above problem we 
have, 

100 X 40 = 4000 total watts used, 

4000 

and = 5.36 horse power. , 

764 (watts per H. P.) 

148 



How many 16 c. p. 3.1- watts (per candle power) 
could be run from a 25 K. W. (kilowatt) dynamo? 
IK. W. = 1000 watts and 25 X 1000 == 25000 watts, 
then 16 X 3.1 = 49.6 watts for each lamp, 
and 25000 -f- 49.6 = 504 lamps at dynamo. 
To find size of wire to use on two wire service, the 
following is given as a formula: 

CXDX21 

M = 

P 

In this M = circular mils of wire which is found 
by squaring the diameter of any wire expressed in 
mils (1 mil = .001"). 

C = Current amperes, 

D = Average distance one way of wire, 

P = Per cent loss allowed, 

21 = Constant. 

Example : Find size of wire required to carry 
current for 150, 16 c. p. 3.5 watt lamps at 120 volts 
from a dynamo a distance of 500 feet at a loss of 5%. 

120 x -05 = 6 volts loss ; 

so, 120 — 6 = 114 volts applied to lamps ; 

then 150 X 16 X 3 - 5 = 84 00 watts required, 

and 8400 -f- 114 volts =73.6 amperes current. 

Then using formula as given above for size of 
wire we have : 
73.6 (current) X 500 (distance) X 21 (constant) 

5 (per cent loss) 
154560 mils size of wire. 

Looking at the table we find this area to be closest 
to a No. 000 wire. 



149 



GEARS 

In finding the relative speeds of two gears when 
in mesh, no matter whether they be straight, spur or 
bevel gears, we proceed as follows : 

number teeth X speed = number teeth X speed, 
(one gear) (other gear) 

From this we can find the number of teeth or the 
speed of either gear when the other data is given. 

Examples : If a 30 teeth gear is running 210 
r. p. m. and is meshing with a gear of 50 teeth, what 
speed is the latter gear making? 

30 X 210 = 50 X (X) ; 
then (X) speed of 50 t gear = 

30 X 210 

= 126 r. p. m. 

50 

A 40 teeth gear is running 200 r. p. m. and is do- 
sired to run another gear at 250 r. p. m. ; how many 
teeth must the latter gear have? 

40 X 200 = (X) X 250 ; 
then, (X), No. teeth in gear = 

40 X 200 

= 32 teeth. 

250 

When two gears are meshing the speed of one 
gear is opposite to that of the other. 



150 





Thus in Fig. 5 the two gears A and B are running 
.toward each other, and the two gears,. C and D, are 
running away from each other. 

When two gears are connected by a carrier gear 
as shown in Fig. 6 the direction of speed of the two 
gears considered are in the same direction. Also the 
relative speeds of the two gears are not affected. 




J 



A 



® 

As shown in (a) Fig. 6 the directions of A and C 
are both the same when connected by the carrier 
gear, B, neither is the relative speeds of the two 
gears, A and C, affected by this carrier gear. Sup- 
pose A in (b) is on a shaft making 100 r. p. m. and 
has 40 teeth, it drives through the carrier gear, B, 
to C which has 60 teeth, what is the speed of this 
roll? 

40 X 100 = 60 X (X) ; 
then (X) speed of roll = 



40 X 100 
60 



= 66.6 r. p. m. 



151 



Yarn Calculation 



As all calculations in the mill are based on yarn 
and hank roving numbers, it is important to have a 
thorough knowledge of the methods used to determ- 
ine these numbers. As a general rule yarn numbers 
are in an inverse ratio to the sizes or diameter of 
yarns. Thus a number 1 yarn has a greater diam- 
eter than a number 2 yarn. In like manner a num- 
ber 16 yarn is larger in diameter than a number 30 
yarn. 

r-The number of a yarn is based on length and 
weight. If we know the weight, either in pounds, 
ounces or grains, of a certain length of yarn we 
can then determine its number. The standard used 
in finding yarn numbers is based on the method 
which has come into constant universal use. This 
standard first takes 840 yards as a length and is 
called one "hank." It also takes 1 pound as the 
weight of a hank if it be number 1 yarn. Then for 
one hank of 840 yards if the weight is reduced the 
yarn becomes smaller and the number is increased. 
Thus if a hank (840 yards) weighs only % pound 
the number would be increased twice and we call it 
a. number 2 yarn. If the weight be reduced to % 
pound the number would be increased to 3 and so 
on. Again,, if we increase the length of a certain 
weight the yarn will become smaller in diameter and 
the number then increased. For instance, if we in- 
crease the length of 1 hank (840 yards) so as to 
make 2 hanks (1680 yards) and if it still weighs 1. 



,152 



pound it. is called. a.. number 2 hank or a number 2 
yarn. From this we see then that as the number of 
hanks in 1 pound is increased, the number is in- 
creased. If a pound of yarn contains 6 hanks 
(= 840 X 6 = 5040 yards) it is a number 6 yarn 
or hank roving. If a pound contains 10 hanks 
(= 840 X 10 = 8400 yards) it is a number 10 hank 
•roving or yarn. In like manner in one pound of 
yarn we have the following : 

1 hank (840 yards) =No.l hank, roving or yarn, 

2 hanks (840X2=1680) =No.2 hank, roving or yarn, 

3 hanks (840x3=2520)=No.3 hank, roving or yarn, 

4 hanks (840X4=3360) =No.4 hank, roving or yarn, 
and so on. 

The number of hanks., then, in 1 pound is the 
number of the roving or yarn. 

Examples: 1. How many hanks in 1 lb. of No. 
30- yarn? 

2. How many hanks in 2 lbs. of No. 40 yarn? 

3. If a yarn is No. 18.25, how many hanks in 1 lb. ? 

4. If 1 lb. of roving has 2.75 hanks, what is the 
No.? ■ 

5. How many yards in 1 lb. of No. 3 hank roving? 

840 = 1 hank; 
then 840 X 3 = 2520 yds. in 1 lb. 

6. How many yards in 2 lbs. of No. 2 hank roving? 

1 lb. of a No. 1 hank = 840 yds. ; 

2 lbs. = 2 X 840 — 1680 yds. in 1 lb. of No. 1 hank; 

then 1680 X 2 = 3360 yds. in 2 lbs. of 

No. 2 hank roving. 



ms 



7. How many yds. in 1 lb. of No. 20 yarn? 

No. 1 yarn = 840 yds. ; 
then No. 20 = 840 X 20 = 16800 yds. 

8. How many yds. in 1 lb. of No. 30 yarn? 

No. 1 yarn = 840 yds. ; 
then No. 30 = 840 X 30 = 25200 yds. 

9. If a No. 1 hank roving has 840 yds., what is its 
weight ? 

1 lb. = 840 yds. = No. 1, 
then 840 yds. of No. 1 hank roving weighs 1 pound. 

10. How many pounds in 1680 yds. of No. 1 hank 
roving ? 

1 lb. = 840 yds. ; 
then 1680 yds. = 1680 / 840 = 2 lbs. 

11. How many pounds in 16800 yards of No. 20 
yarn ? 

1 lb. of No. 20 = 840 X 20 = 16800 yards ; 
then 16800 yds. = 16800 /i 6 800 = 1 pound. 

12. How many pounds in 33600 yards of No. 10 

yarn? 

1 lb. No. 10 = 840 X 10 = 8400 ; 
then 3360 %4oo = 4 pounds. 

13. If 2520 yards of roving weigh 1 pound, what 

is its No. ? 

1 lb. of No. 1 = 840 yds. ; 
then 1 lb. of (X) yarn = 252 % 40 = No. 3 roving. 



154 



14. If 840 yards of roving weigh 2 pounds, what 
is its No. 1 

1 lb. of No. 1 = 840 yds. ; 

2 lbs. of No. 1 = 840 X 2 = 1680 yds. ; 

then if 2 lbs. of No. 1 = 1680 yds., 2 pounds of 

(X) yarn = ^% 680 = No.. % = .5 H. R. 

. From the above examples we see that if we have 
the pounds and yards given we can find the num- 
ber. Thus if we have 1680 yards weighing 1 pound, 
the number is found by dividing 1680 by the No. 
yards of No. 1 in 1 pound which is 840. Then 
1680 -r- 840 X 1 = No. 2. Again if we have 25200 
yards weighing 3 pounds we find the number thus, 

25200 
= No. 10. 



840X3 

(a) Then if we have weight in pounds and yards 
given we have, 

number yards 

(1) = No. 

840 X pounds 

Also if we have the number and the number of 
yards given we have, 

number yards 
(2) = Pounds. 

840 X No. 

Again if we have pounds and No. of yarn or rov- 
ing given, we can find the number of yards thus, 

(3) 840 X No. X pounds = number of yards. 



155 



(b) If the length is given in yards and the weight 
in ounces, it is only necessary to find, first, the num- 
ber of yards in 1 ounce of No. 1 hank. If there are 
840 yards in 1 lb. = 16 ounces of No. 1, then there 
would be 840 -=- 16 = 52.5 yards of No. 1 in 1 
ounce. Then as in (a) we will have, 

number yards 

(1) = No. ' 

52.5 X ounces 

number yards 

(2) = ounces. 

52.5 X No. 

(3) 52.5 X No. X ounces = number yards. 

(c) If the length is given in yards and the weight 
in grains we must likewise find the number of yards 
of No. 1 in 1 grain. If there are 840 yards of No. 1 
in 1 lb. = 7000 grains, then there would be 840 -=- 
7000 = 21 / 175 = .12 yards of No. 1 in 1 grain. Then 
as in (b) we have,, 

number yards 

(1) = No. . 

.12 X grains 

number yards 

(2) = grains. 

.12 X No. 

(3) .12 X No. X grains = number yards. 

CARD ROOM PROBLEMS 

In finding the H. K. No. (hank roving No.) of 
caret or drawing slivers the; usual practice is to 
weigh 1 yard in grains. If it is desired to find th3 



156 



H. R. No. this weight in grains is multiplied by .12 
and divided into the number of yards, 1. As 1 yard 
is always taken a constant is found which will give 
the H. R. No. by simply dividing the weight in 
grains into this constant. This constant is as fol- 
lows : 

number yards 

If = No. it is the same as 

.12 X grains 

number yards 

1_ grains. 

.12 

As 1 yard is always taken then we have 

1 1 1 12 

'— grains, now = '— ■ — — because 

.12 .12 1 100 

.12 = i2/ 10 o, and 1 = ft 
Then 1 / 1 ~ 12 Aoo = 100 /i2 = 8% ; then we have, 

8i/ 3 (= 8.333) 
= No. if 1 yard is taken. 



grains 

In finding the H. R. No. of roving 12 yards are 
usually taken and weighed in grains. We also find 
a constant which by dividing the weight in grains 
into it the H. R. is found. This constant is found 
as follows : 

number yards 

If = No., it is the same as 

.12 X grains 

number yards 

-f- grains. 

.12 

157 



12 

And as = 100 we have, 

.12 

100 
= No. if 12 yards are taken, 



grains 

100 

or = grains. 

No. 

Examples : 1. If 12 yards of roving weigh 25 
grains, what is the H. R. No. ? 

100 -=- 25 = 4 H. R. 

2. 12 yards of roving weigh 35 grains ; what is 
the H. R. No. ? 

100 -=- 35 = 2.857 H. R. 

3. How many grains ought 12 yards of 1.75 H. R. 
to weigh? 

100 -^ 1.75 = 57.14 grains. 

4. Four 12-yard lengths of roving are taken 
weighing 52.5, 51.75, 52 and 52.25 grains ; what is 
the average H. R. No.? 

An average weight is found by adding all weights 
and dividing by the number of weighings made. 
Thus, 

52.5 

51.75 

52. 

52.25 

4 ) 208.50 

52.12 average weight. 

100 
= 1.91 H. R. 



52.12 

158 



5. What H. R. is a 56 grain drawing sliver? 

1 yard = 56 grains, 
12 yards = 12 X 56 = 672 grains ; 
then 100 -f- 672 = .1488 H. R. ; 
or 8.3333 -=- 56 = .1488 H. R. 

6. If a 48 grain drawing sliver is fed to the slub- 
ber and 1 yard of sliver makes 3 yards of roving, 
uwhat is the H. R. being made on slubber? 

1 yard roving would weigh 48 -~ 3 = 16 grains ; 

then 12 yards would weigh 12 X 16 = 192 grains, 

and 100 / 192 = .52 H. R. 

7. If hank clock on a frame registers 8 hanks run 
in a day and 4 H. R. is being made, what is the 
pounds production per spindle per day? 

8 hanks = 840 X 8 = 6720 yards per day. 
From 2 under (a). 

6720 

2 pounds, 



840 X4 

or as 6720 = 840 X 8 we have, 

X8 8 

= — = 2 pounds per spindle per day. 



X4 4 

From this we get the rule for pounds production. 

No. hanks from clock X No. spindles 

= pounds 

H. R. production. 

SPINNING ROOM PROBLEMS 

1. If 12 yards of H. R. weigh 40 grains, what is 
its No.? 

159 



From (1) in (c) we have 

12 100 
= = 2.5 H. R. No. 



.12 X 40 40 

2. If 12 yards of 2.5 H. R. are fed into spinning 
frame, one "end up," and 120 yards are delivered, 
what is the yarn No.? 

From (1) under (c) we have 

12 100 

= = 40 grains, 

.12 X 2.5 2.5 weight of 12 yds. 

of 2.5 H.R. 

Then if 120 yards of yarn are made from this it 
will weigh approximately 40 grains. Then we have, 

120 1000 

= 25 yarn No. 



.12 X 40 40 

The usual practice in finding the number of yarn 
is to reel off 120 yards and weigh in grains. The 
weight of this 120 yards in grains divided into the 
constant will give the number of yarn. 

If 120 yards are taken Ave have, 

120 120 

-r- grains, from (1). under (c), and — — = 1000 



.12 .12 

which is our constant. Then we always have, 

1000 

= yarn No. 



wt. in grains of 120 yds. 



3. If 120 yards of yarn weigh 25 grains, what is 
the number? 



160 



1000 

= 40 yarn No. 

25 

4. If a 1 inch front roll is running 136 revolutions 
per minute on 30 's yarn, what is the pounds pro- 
duction in a day per spindle ? 

Circumference of roll = 1 X 3.1416 = 3.1416 
m inches, and 3.1416 X 136 = 427.2576 

inches delivered in 1 minute ; 

427.2576 -=- 36 = 11.86 yards delivered in 1 minute ; 

60 X 10 hours = 600 minutes in a day ; then 

11.86 X 600 = 7116.00 yards per day of 

10 hours, and from (2) under (a) 

7116 7116 

= .282 lbs. per spindle 



840 X 30 25200 per day. 

5. If a spool has 14 ounces of No. 28 's yarn on it, 
what is the length? 

From (3) under (b) we have, 
52.5 X 28 X 14 = 14700 yards. 

6. Spools are to be made with enough 30 's yarn 
to make 18000 yard warp on warper ; what should 
be weight in ounces of yarn on each spool? 

From (2) under (b) we have, 

18000 

= 11.4 ounces or 12. 

52.5 X 30 

7. If a warper beam has 436 ends of No. 30 's and 
is 16000 yards long, what should it weigh in pounds ? 

436 X 16000 = 6,976,000 yards of yarn on beam ; 
then from (2) under (a) we have, 



161 



6,976,000 

= 276.8 pounds. 

840 X 30 

8. A warper beam weighs 400 pounds and has 
342 ends of 28 's on it; what is the length of warp? 

400 -=- 342 = 1.169 pounds, weight of each end on 

beam; then from (3) under (a) we have, 

840 X 1-169 X 28 = 27494 yards. 

WEAVE ROOM PROBLEMS 

1. 8% size is put on a warp of 30 's which has 
2040 ends. If the beam has 15 cuts, each 64 yards 
in length, what is the weight of yarn on beam? 

15 X 64 = 960 yards, length of warp ; 
then from (2) under (a) we have, 

960 X 2040 

= 77.71 pounds without size. 

840 X- 30 

Now there are two ways of figuring the % of the 
size. First, it may be figured based on the result- 
ing weight of finished warp; or, second, it may be 
figured on the weight of yarn before sizing. To 
illustrate the first case,, if we have 2000 pounds of 
yarn going to slasher and the finished yarn on beams 
weighs 2200 pounds, we have, 

2200 — 2000 = 200 pounds of size put on, and 

200 X 100 

= 9.09%. 

2200 

To illustrate the second case we can take the same 
example and we have, 



162 



2200 — 2000 = 200 pounds of size put on, and 

200 X 100 

= 10%. 

2000 

It seems that as the size is put on yarn and sold 
as part of the finished cloth the resulting weight of 
^yarn after sizing should be the basis from which to 
figure the % of size. 

Then if we use the first method of figuring % in 
example 1, 77.71 pounds would be 1.00 — .08 = 92% 
of the weight of yarn on beam. Then if 1 pound of 
sized yarn has .92 pounds of cotton in it, (X) pounds 
would have 77.17. This can be solved by the fol- 



lowing 


proportion : 








1 : 


.92 : : X 


: 77.17; 




then X = 


77.17 


= 83.8 pounds. 



.92 

2. A cloth is to be 36 inches wide with 68 ends 
per inch and to be made into cuts of 60 yards. If 
we allow 7% take-up in weaving, how many pounds 
of warp would there be in a cut of cloth using 28 
warp with 8% size? 

36 X 68 = 2448 ends in warp, 

1.00 — .07 = .93 ; 

then 60 yard cut of cloth is 93% of the length 

in a cut of warp. Hence we have, 
60 -r- .93 = 64.5 yards in 1 cut of warp or 64. 

Then from (2) under (a) we have, 

64 X 2448 

= 6.66 pounds of unsized yarn. 

840 X 28 

163 



As under example 1 we have, 

6.66 -f- .92 (1.00 — .08) = 7.239 pounds of 
sized yarn in 1 cut of cloth. 

3. If the cloth in example 2 has 64 picks per inch, 
how many wards of filling would there be in a cut 
of cloth, if the warp is 38 inches at the reed? 

38X64X^0 

= 2432 yards of filling in 1 

fifi yard of cloth. 

then 60 X 2432 = yards of filling in 1 cut. 

If the cloth is to weigh 4 yards per pound, what 
must be the filling No. ? 

4 yards per pound = 6 % = 15 pounds per cut. 

From example 2 we see that there are 7.239 
pounds of warp in 1 cut. Then 15 — 7.239 = 7.761 
pounds of filling in 1 cut. 

From (1) under (a) we have, 

60 X 2432 
= 22.38 filling No. 



840 X 7.761 

PLY YARNS 

When two single yarns are twisted together to 
form a strand, the resulting yarn is said to be "two 
ply yarn." When 3 single yarns are twisted to- 
gether the result is said to be "three ply yarn,"- 
and so on for as many single strands as are twisted 
together. Twisted yarns are shown in figures by 
placing the number which represents the single 
yarn above a line and the figure Avhich shows how 



164 



many of such strands were used below this line. 
Thus, a 4 % yarn means that 2 ends of single 40 's 
are twisted together, 3 % two ends of 30 's, and so on. 
When two ends of single yarn are thus twisted to- 
gether the resulting hank number is practically 
one-half of the single yarn. It will be a little 
heavier owing to the contraction in twist. Thus a 
3 % yarn = 15 counts, a 4 % = 20 counts, a 3 % = 10 
*counts, and so on. 

When two yarns of different sizes are twisted 
together the resulting number is found by multiply- 
ing the two single numbers together and dividing 
this product by the sum of the two numbers. 

Example : If a 30 and 40 are twisted together 
what is the resulting number? 

30 X 40 1200 
= = 17.14. 

30 -f 40 70 

Proof : 120 yds. 30 's = 1000 -=- 30 = 33% grains. 
: 120 yds. 40 's = 1000 -*- 40 = 25 grains. 

120 yds. twisted yarn 58% grains. 

Then 1000 -^ 58.33 = 17.14. 

WEIGHT OF YARN ON BEAMS, SPOOLS, ETC. 

Sometimes it becomes necessary to find the weight 
of yarn contained on a beam or spool of certain 
size ! To do this 60 cubic inches of yarn are allowed 
to 1 pound. 

Example : A warper beam is 54 inches between 
heads and has a 9 inch barrel and 24 inch heads; 
how many pounds of yarn will it hold? 

First, multiply the result of adding the diameters 

165 



of head and barrel by the result of subtracting bar- 
rel from head. 

24 + 9 = 33 and 24 — 9 = 15 ; 
then 33 X 15 = 495. 

Second, multiply the above result by .7854, 

then .7854 X 495 = 388.773 square inches. 
Third, multiply this area by distance between 
heads, 

then 388.773 X 54 = 20993.742 cubic 
inches capacity. 

Fourth, as 60 cubic inches = 1 pound, then 
20993.742 -=- 60 = 349.8 pounds. 



166 



Draft Calculations 



The process of drawing cotton sliver and roving 
by means of rolls was invented by John Wyatt in 
1737. This process was afterwards improved upon 
J)y James Hargeaves, 1764, Arkwright in 1767 and 
Samuel Compton in 1779. Arkwright is usually 
given the credit for this improvement in cotton 
spinning. 

The drawing out of slivers or roving is done by 
feeding into a roller with a constant speed a certain 
length of sliver or roving. The cotton then passes 
to another roller which has a faster speed than the 
previous one, thus drawing out the strand of cotton 
between the two. This drawing out of strands com- 
posed of cotton fibers can be done to a greater ex- 
tent from fine strands than from heavier ones. For 
this reason we see greater drawing as the strand 
gets finer in the mill. For instance, starting with 
the first process in drawing we gradually increase 
this process to the last machine. 

This drawing out of a strand of cotton is com- 
monly called Draft. Draft then means the amount 
of attenuation or drawing out that this strand re- 
ceives in a certain machine. If a yard of sliver is 
fed into, a machine and four yards delivered the re- 
sulting strand is one-fourth as large as it was at the 
beginning,, and draft is said to be 4. If 2 yards are 
fed into a set of rolls and 8 yards are delivered the 
ratio of the fed strand to delivered strand is 1 to 
4 and the draft is 4. From this we find the fol- 
lowing : 

167 



EFFECT OF DRAFT ON LENGTH 

If 2 yards fed = 8 yards delivered, draft is 4. 
If 4 yards fed = 20 yards delivered, draft is 5. 
If 400 inches fed = 2400 delivered draft is 6. 

From this we see that, 

(a) length delivered -=- length fed = draft. 

Also if 3 yards are fed and draft is 4 = 12 yards 
delivered. 

If 4 yards are fed and draft is 5 = 20 yards de- 
livered. 

If 20 inches are fed and draft is 4 = 80 inches 
delivered. 

From this we see that, 

(b) length fed X draft = length delivered. 

Also if 36 yards are delivered and draft is 6 = 6 
yards fed. 

If 100 yards are delivered and draft is 10 = 10 
yards fed. 

If 3600 inches are delivered and draft is 12 = 300 
inches fed. 

From this we again see that, 

(c) length delivered -4- draft = length fed. 

EFFECT OF DRAFT ON WEIGHT 

In considering the weights of certain sliver or 
rovings the effect of draft is as" follows: 

If one yard weighing 5400 grains is fed, and 1 
yard delivered weighs 50 grains, the draft is 90. 

If 12 yards weighing 25 grains are fed and 12 
yards delivered weigh 10 grains- the draft = 2.5. 



168 



(a) From this we see that, grains in 1 yard fed -r- 
grains in 1 yard delivered = draft. Or grains in 12 
yards fed -=- grains in 12 yards delivered = draft. 

Also if 1 yard weighing 50 grains is delivered and 
draft is 100 the weight of 1 yard fed is 5000 grains. 

If 12 yards weigh 25 grains delivered and draft 
is 4 the weight of 12 yards fed is 100 grains. 

(b) From this we see that, grains in 1 yard de- 
livered X draft = grains in 1 yard fed. Or grains 
in 12 yards delivered X draft = grains in 12 yards 
fed. 

Also if 1 yard weighing 4500 grains is fed and 
draft is 90, the weight of 1 yard delivered is 50 
grains. 

If 12 .yards weighing 100 grains is fed and draft- 
is 4, the weight of 12 yards delivered is 25 grains'. 

(c) From this we see that, grains in 1 yard fed -r- 
draft = grains in 1 yard delivered. Or grains in 
12 yards fed -=- draft = grains in 12 yards delivered. 

EFFECT OF DRAFT ON WEIGHT WHEN 
DOUBLING 

If two or more laps, sliver or rovings are fed to 
the rolls it is only necessary to find the resulting 
weight of a certain length after this doubling. Thus 
if 4 laps are fed into a finisher picker and each lap 
weighs 14 ounces to the yard the resulting weight of 
1 yard being fed is 4 X 14 = 54 ounces. If a lap 
weighing 14 ounces to the yard is delivered from 
this the draft is 4. From this we have, 

No. of doublings X weight of each 

(a) = draft. 

weight delivered 

169 



and 



No. of doublings X weight of each 

(b) = weight 

draft delivered. 

or 

weight delivered X draft 

(c) = weight fed. 

No. doublings 

EFFECT OF DRAFT ON H. R. NOS. AND 
YARN NOS. 

In considering the effect of draft on numbers of 
yarns or hank roving it is only necessary to remem- 
ber that the size of a yarn or roving is reduced as 
the number is increased, or that the yardage is in- 
creased as the number is increased. Thus a No. 1 
yarn or hank roving has 840 yards in 1 lb.„ a No. 2 
yarn or hank roving has 1680 yards in 1 lb., and 
so on. 

Then as draft deals directly with relative lengths 
fed and delivered it has a similar effect on numbers 
of yarns and hank roving fed and delivered. 

If a 1 hank roving is fed and = 2 H. E. delivered, 
draft is 2. 

If a 2 hank roving is fed and = 8 H. B. delivered, 
draft is 4. 

If a 6 hank roving is fed and = 36 yarn delivered, 
draft is 6. 

EFFECT OF DRAFT ON NOS. WHEN 
DOUBLING 

If the H. E. is doubled the resulting H. R. fed is 
found by dividing the No. H. R. by the No. 
doublings. 



170 



Thus if a 2 H. E. is being fed two ends into one, 
the resulting H. R. is 1. If a 4 hank roving is being 
fed two ends into one, the resulting H. R. is 2, etc. 

Then, 

If a 2 H. R. is fed doubled 2, and = 2 H. R. de- 
livered, draft = 2. 

If a 4 H. R. is fed doubled 2, and = 8 H. R. de- 
livered, draft = 4. 

If a 6 H. R. is fed doubled 2, and = 30 yarn de- 
livered, draft = 10. 

From this we see that, 

H.R. or yarn delivered X No. doublings 

(a) = 

H.R. fed draft 

Also, if 2 H. R. is fed doubled 2, and draft is 
2 = 2 H. R. delivered. 

If 4 H. R. is fed doubled 2, and draft is 4 = 8 H. 
R. delivered. 

If 6 H. R. is fed doubled 2, and draft is 10 = 30 
yarn delivered. 

From this we see that, 

H.R. fed X draft 

(b) = No. H.R. or yarn 

No. doublings delivered. 

When Doubling" Is 2 

If 2 H. R, is delivered and draft is 2 = 2 H. R. 
fed (single). 

If 8 H. R. is delivered and draft is 4 = 4 H. R. 
fed (single). 

If 30 yarn is delivered and draft is 10 = 6 H. R. 
fed (single). 

From this we see that, 



171 



H.R. or yarn No. delivered X No. doubling 

(c) : = 

draft H. R, 

(single) fed. 

DRAFT OF MACHINE 

The draft in a machine is figured from the rel- 
ative length of sliver or roving being received by 
one roll and that delivered by another. 

For instance,, if back roll takes in 300 inches of 
sliver or roving in one minute and front roll de- 
livers 900 inches in the same time, the front roll will 
deliver 3 inches while the back roll is taking in one 
inch. Then the draft is said to be 3. 

Example : If back roll is 1%" in diameter and is 
making 100 revolutions per minute, and front roll 
is 1%" in diameter and makes 200 revolutions per 
minute, what is the draft? 

The number of inches per minute taken in by the 
back roll Avill be the same as the surface speed of 
this roll expressed in inches per minute. The sur- 
face speed is found by multiplying together the di- 
ameter, iy 8 " (= 1.125"), 3.1416 and its speed, 100. 

Then, 1.125 X 3.1416 X 100 = 353.43 inches taken 
in by back roll in 1 minute. 

Also, 1%" (= 1.375"), X 3.1416 X 200 = 863.94 
inches delivered by front roll in one minute. 

Then the draft is found by dividing 863.94 by 
353.43 as seen in (a) under "Effect of Draft on 
Lengths. ' ' 

Then 863.94 -=- 353.43 = 2.44 draft. 

863.94 

If (1) = draft, 

353.43 



172 



1.375 X 3.1416 X 200 

then (2) = draft, 

1.125 X 3.1416 X 100 

because (1) and (2) are equal or (1) is 
the same as (2). 

By cancelling the 3.1416 on the top and bottom 
we have left, 

2 
1.375 (front roll dia.) X 200 (speed front roll) 

(a) ^ = 

1.125 (back roll dia.) X W (speed back roll) 

2.750 

= 2.44 draft, or Ave may show this in the 

1.125 following manner : 

1.375 X 200 
= 2.44 draft. 



1.125 X 100 



200 

If we consider the above, , as finding the rel- 

100 

ative speeds of the two rolls, as it is, it is not nec- 
essary to know the speed of either roll to determine 
the draft. All we want to know is to find out ratio 
of speed between the two,, or to find when the back 
roll makes 1 turn how many turns the front roll will 
make. This can be found out by considering the 
gearing between the two. 

If we consider the following sketch as the gear- 
ing between front roll and back the total draft be- 
tween the two is found as follows : 



173 



I '/a" BACK ROLL 



_/~ 



24 B 



A-70 



l^-fRONT ROLL 






FIG. 7. 





C-96 








D-30 



If we consider the back roll as making 1 turn 
then the gear, A, will make 1 turn and the gear, B, 
will make 



1X70 



24 



turns = 2.916 -j- turns. 



Then, if B makes 2.916 turns the gear, C, on same 
stud will make 2.916 turns and the gear, D, will 
make 



2.916 X 96 



= 9.331 turns. 



30 



From equations we saw that if we use 2.916 in 
the last process we could substitute for it the value 

1X70 



24 

and bring all of the calculation under one head. 



174 



1 X 70 96 70 X 96 

Then (which is 2.916) X — = 



24 . 30 24 X 30 

and using cancellation we have, 
4 

W X H 28 

= — = 9.333 turns of front roll while 

4fy£ X fift 3 back roll turns 1 time. 

3 ft 

If then we consider the gears B (= 24 teeth) and 
D {== 30 teeth) as the drivers and the gears C 
(= 96 teeth) and A (= 70 s teeth) as driven gears, 
then the relative speed of the back to the front roll 
is found by multiplying the driven gears together 
and dividing this product by the driving gears mul- 
tiplied together. 

Now, as we saw under (a) that, 

diam. of front roll X speed of front roll 

= draft. 

diam. of back roll X speed of back roll 

And as in this example the front roll is 1% inches 
in diameter = 1.375, and back roll 1% inches in 
diameter = 1.125 we have, 

1.375 X 9.333 12.832 

(b) = = 11.4 draft. 

1.125 X 1 1.125 

Again, if in the above equation, (b) 
9.333 70 X 96 



1 24 X 30 

we can substitute and have the following 

175 



1.375 X 70 X 96 

(c) = 11.4 draft. 

1.125 X 24 X 30 

RULE TO FIND DRAFT 

In order then to find the draft of any machine, 
start with the diameter of back roll, or first roll 
into which cotton is fed. Place this diameter under 
the line as seen above and place the diameter of 
front roll,, or last roll through which cotton passes, 
on top of the line. Then place all driven gears 
above line with the sign, X, between them. Then 
place all driver gears below the line with the sign, 
X, between them and solve by cancellation. 

CHANGE GEARS 

In Fig. 7 if the gear, B, is a change gear the draft 
is affected by changing the number of teeth in this 
gear. If the number of teeth is increased the back 
roll is run faster and the draft is reduced. 

Also if the number of teeth is decreased the back 
roll runs slower and the draft is increased. From 
this we see that a larger gear produces less draft 
and a smaller gear produces more draft. Then the 
draft is inversely proportional to the number of 
teeth in change gear. Thus if a 24 teeth gear pro- 
duces a draft of 11, what gear will produce a draft 
of 6 ? If we use a direct proportion to. find this we 
have, 

24 : 11 : : X : 6 

6X24- 

then = 13 teeth gear, 

11 



176 



which being a smaller gear than the 24 teeth ought 
to produce more draft. 

Hence this proportion is not true, because we want 
less draft than that produced with the 24 teeth 
gear. Then if we turn the proportion completely 
around and say 24 teeth is to 6 draft as X teeth is 
to 11 draft w T e have an inverse proportion as follows : 

24 : 6 : : X : 11 ; 

11 X24 

then = 44 teeth change gear. 

6 

The above proportion is correct and may be used 
to advantage. 

DRAFT CONSTANTS 

As the gear, B, in Fig. 7 can be changed to pro- 
duce different drafts, the number, 24, could be left 
out in the equation (c), and the resulting number 
thus obtained when divided by whatever number 
draft gear is to be used will give the draft. 

Thus the draft constant in (c) will be, 

7 32 
1.375 X JT0 X H 

= 273.77 constant. 

1.125 X (X) X^ 

Now as the number 24 was left out below the line 
if this number be divided into 273.77 the draft will 
be obtained. 

Thus, 273.77 -=- 24 = 11.4 draft. 

Such constants are usually obtained for all ma- 
chines and when it is desired to change the draft it 
is quickly obtained. 

177 



If the constant 273.77,, when divided by the gear, 
24, gives the draft, 11.4, then the draft, 11.4, divided 
into the same constant, 273.77, will produce the gear, 
24, as follows : 

273.77 -^ 24 gear = 11.4, 
and 273.77 -=- 11.4 draft = 24 gear. 

Then when we have a draft constant we can ob- 
tain the draft gear to use for any draft, and also 
the draft obtained by using any draft gear. 

constant 

Thus, = draft. 

change gear 

constant 

= change gear. 



draft required 



17P 



DRAFT OF A PICKER 























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179 



Using the above sketch the draft of a finisher or 
intermediate picker is found as follows : 

cal. 

roll ACBGI KMO 

9 X32X22X90X12X28 X20X15X13 

= 3.8 draft. 

2.5X40X 9X1 X24X45 X52X72X54 
feed B D F H J L N P 
roll change 

If the change gear, J, is left out Ave have the 
constant number, 171. If this gear, 45, be divided 
into 171 we get 3.8 draft. 

constant 

Then = draft. 

change gear 

constant 

And = change gear. 

draft 

Examples : If 4 laps weighing 14 ounces to the 
yard each are fed to this machine using a 45 teeth 
gear, what will each yard of lap weigh in front? 

constant 171 
= 3.8 draft. 



change gear 45 
Then as 
number of doublings X wt. of each 
draft 



= wt. in front. 



From (b) under "Effect of Draft on Weight 
When Doubling." 



180 



4X14 

"We have = 14.73 ounces, weight of each 

3.8 yard delivered. 

As there is some of this lost in going through in 
the form of waste this allowance must be made. 
Then allowing 2% for waste we have 14.73 ounces 
as 100% of weight fed in. 

Then, 14.73 X -02 = .29 ounces lost as waste. 

Then, finished lap should weigh 

14.73 — .29 = 14.44 ounces. 

To find the % of waste made. 

If 4 laps each weighing 14 ounces to the yard are 
fed and 1 yard of finished lap weighs 14% ounces 
and draft is 3.8, the finished lap should weigh 

4X14 

— = 14.73 ounces. 

3.8 

But if it weighs only 14% ounces the loss is 
14.73 — 14.5 = .23 ounces. Then the % of waste 
would be 

.23 X 100 

= 1.56% waste. 

14.73 

Example : If 4 laps each weighing 15 ounces are 
fed to machine and the lap to be delivered is to 
weigh 14% ounces, what draft gear must be used, 
allowing 2% for waste? 

4 X 15 = 60 ounces, weight of 1 yard 

fed to machine. * 



181 



As 14.75 ounces is to be the finished lap after al- 
lowing for waste, the theoretical weight would be 
found as follows : 

14.75 is (1.00 — .02) 98% of the theoretical weight; 
then 14.75 -=- .98 = 15.05, weight of 1 

yard if no waste were made. 
Then 60 -^ 15.05 = 4 draft practically. 

Using the constant, 171, as found above we have, 

171 constant 

= 42.7 or 43 draft gear. 



4 draft 

Example : If a 14 ounce lap is required and draft 
in machine is 4, what must be the weight per yard 
of each of the 4 laps if 2% waste is allowed? 
14 ounces is (1.00 — .02) 98% of the 
theoretical weight allowing for waste. 
Then 14 -=- .98 = 14.28 ounces lap if no waste 
were made. 

Then if draft is 4, 14.28 X 4 = 57.12 ounces, 
weight of 1 yard of cotton going in feed roll. 
As there are 4 laps going in, 
57.12 -f- 4= 14.28 weight of each lap behind. 



182 



DRAFT OF A CARD 




FLOOR PLAN OF 40-INCH CARD WITH 27-INCH DOFFER 

FIG. 9 

183 



Using the above sketch the draft of a card is 
found as follows. 

coiler 
roll 
2 X 120 X 40 X 214 X 27 

= 76.73 draft. 

2.25 X 20 X 45 X 21 X 17 
feed bevel 
roll change 

If the change gear, 20, is left out we will get a 
result of 1534.6 which is the draft constant. 

If 1534.6 is divided by 20 the draft, 76.73, is found 

thus, 

1534.6 constant 

= 76.73 draft. 

20 change gear 

Or, 

1534.6 constant 

= 20 change gear. 

76.73 draft 

Example : If a 12 ounce lap is fed to the card 
using an 18 teeth change gear, what will be weight 
of sliver delivered? 

constant 1534.6 

= 85.2 draft with 18 

change gear 18 teeth gear. 

A 12 ounce lap = 12 X 437.5 = 5280 grains per 
yard; then, 

5280 

= 61.9 grain sliver delivered. 

85.2 draft 



184 



As there is some waste the resulting sliver will 
not weigh quite this much. If 4% is allowed for 
Avaste this weight would be reduced as follows : 

61.9 = 100% stock fed; 

so 61.9 .X -04 ===== 2.476 grains lost in waste ; 

and 61.9 — 2.476 = 59.424 grains 

actual weight delivered. 

* To find actual % of waste in card. 

Example : If the weight of lap is 12 ounces == 
5280 grains and card has a draft of 90 and is de- 
livering a 56 grain sliver we have, 

5280 -f- 90 1 ===== 58.66 grain sliver, theoretical. But as 

it only weighs 56 grains the loss would be 

58.66 — 56 = 2.66 grains ; then 

2.66 X 100 

= 4.5+% waste. 

58.66 

Example : If 12 ounce laps are used and a 54 
grain sliver is to be produced, what draft gear must 
be used? Allow 4% waste. 

A 54 grain sliver would be (1.00 — .04) 96% of 
total weight that should be delivered if there were 
no waste. 

Then, 54 -f- .96 = 56.25 theoretical weight of sliver. 

A 12 ounce lap = 5280 grains; 

then 5280 -f- 56.25 = 93.86 draft required 

for 54 sliver ; and, 

constant 1534.6 

= 16-}- draft gear. 

draft 93.86 



185 



Example : If a 52 grain sliver is desired and the 
card has a draft of 95, what must the lap weigh in 
ounces. Allow 4% waste. 

52 -+- .96 (1.00 — .04) = 54.16 grain sliver 

if no waste is made ; 

then 54.16 X 95 draft = 5145 grain lap, 

and 5145 -i- 437.5 grains in 1 ounce = 11.76 

or 11% ounces lap. 

What draft gear will give 95 draft? 

constant 1534.6 

= 16 gear. 

draft 95 

DRAFT OF DRAWING FRAME 



BACK ROLL 



MtGEJ 



2" CAL ROLL 



As most all drawing is now done with metallic 
rolls it is necessary for us to remember the follow- 
ing in figuring the draft : 

Rolls With 16 Pitch 

V/g" diameter figured as 1.66" 
1%" diameter figured as 1.83" 
1%" diameter figured as 2.00" 
1%" diameter figured as 2.13" 



186 



Rolls With 24 Pitch 

1%" diameter figured as 1.50" 
1%" diameter figured as 1.66" 
1%" diameter figured as 1.83" 
V/2" diameter figured as 2.00" 

Using the above sketch the draft is as follows : 
As 1%" back roll = 2" for metallic rolls 

8i 2 
% X 42 X m X W 

= 7.77 draft. 

% X %i X 45 X00 

^ change gear 

If the change gear, 30, is left out of the calcula- 
tion we have the constant,, 7.77 X 30 = 233.10. 

Then if change gear is divided into constant we 
find the draft. Thus, 



and, 



constant 233.10 
change gear 30 

constant 233.10 



7.77 draft; 



= change gear, 30 



draft 7.77 

or, 

draft, 7.77, X change gear, 30, = constant, 233.10. 

Example : If a 54 grain sliver is fed to. drawing 
with doublings of 6, what draft gear would be re- 
quired to make a 52 grain sliver? 

54 X 6 = 324 grains per yard fed in behind ; 
and, 324 -^ 52 = 6.23 draft required. 



187 



constant 233.10 

Then = 37+ teeth change gear. 

draft 6.23 

Example : A 54 grain sliver is required to be de- 
livered with a draft gear of 36. What must be the 
weight per yard of card sliver if the doubling is 6? 



constant 233.10 



6.47 draft. 



draft gear 36 

Then 54 X 6.47 = 349.38 weight of 1 yard fed 
with 6 doublings. 

Therefore, 349.38 -f- 6 = 58.23 grain card sliver. 

Example : If a 52 grain sliver is used behind 
drawing with a 38 teeth change gear and doubling 
6, what will be the weight in grains of sliver de- 
livered ? 



constant 233.10 



6.1 draft. 



draft gear 38 

52 X 6.1 = 317.2 weight of 1 yard fed to machine; 
and 317.2 ~=- 6 = 52.8 grain sliver. 

DRAFT OF A SLUBBER 



BACK ROLL I" DIA. 



30- 

CMAN6E 



FRONT ROLL iVs'DIA- 1.1875 
FIG. 11. 



188 



From the above sketch the draft of a slubber is 
found as follows : 

10 
1.1875 X 56 X 10 
= 6.717 draft. 

1 X 00 X 33 
3 

Leaving out the change gear, 30, Ave have the 
constant number, 30 X 6.717 = 201.51, which can 
be used for finding the draft if gear is given, or for 
finding the gear if draft is given. Thus,. 

constant 

= draft ; 



change gear 

constant 

and = change gear. 

draft 

Example : If a 52 grain sliver is used behind 
slubber with a 42 teeth draft gear, what will be 
H. R. No. in front? 

constant 201.51 

= 4.79 draft. 

change gear 42 

Then if a 52 grain sliver is used behind, the weight 
of 1 yard in front of slubber would be, 52 -r- 4.79 = 
10.85 grains per yard. Then 12 yards would weigh 
12 X 10.85 = 130.2 grains ; then from yarn calcu- 
lations 

100 
= .76 H. E. No. 



130.2 



DRAFT OF AN INTERMEDIATE FRAME 



BACK ROLL I" DIA. 



30— f 

CHANGE 



FRONT ROLL l^'DIA* 1. 1875 
FIG. 11. 



From the above sketch the draft of the interme- 
diate is figured as follows : 

1% 6 ". = 1.1875" 



= 6.71 draft. 



.3958 20 

;j^7ox56x;w> 
1 x w x w 

3.3 $ 
change gear 

If we leave off the change gear, 30, we will have 
6.71 X 30 = 201.3 constant. Then we have, 



and, 



constant 201.3 
change gear 30 

constant 201.3 



= 6.71 draft ; 



draft 6.71 



= 30 change gear. 



Example: If a .62 H. R, comes from slubber and 
a 36 teeth change gear is used on intermediate, what 
will be the H. R. No. in front of intermediate? 



190 



constant 201.3 
= 5.6 draft. 



change gear 36 

As the roving is doubled behind we have from 
<b) under "Effect of Draft on Nos.," 

H. R. fed X draft 

= H. R. No. delivered. 



No. doubling 

Then, 

.62 -f- 2 = .31 H. R. fed ; 
and, 

.31 X 5.6 draft = 1.73 H. R. No. from 
intermediate. 

Example : What change gear should be used to 
make 1.80 H. R. from .62 hank on slubber? 
.62 -f- 2 = .31 resulting H. R. being fed 
after doubling. 
Then from (a) under "Effect of Draft on Nos." 
we have, 

H. R. fed 

H. R. delivered-; = draft, 

No. doubling 
Then, 

.62 1.80 

1.80 -=— = ■ = 5.8 draft ; 

2 .31 
and, 

constant 201.3 

= 34.7 or a 35 teeth gear. 

draft 5.8 

Example : If a 2 H. R. is to come from the inter- 
mediate which has a 28 draft gear, what H. R. must 
the slubber make? 



191 



constant 201.3 



7.19 draft. 



draft gear 28 

Then from (c) under "Effect of Draft on Nos." 
we have, 

H. R. delivered, 2 X doubling, 2 
= H. R. fed (sin- 



gle) = 



draft, 7.19 



= .55 H. R. to be made on slubber. 



7.19 



DRAFT OF FINE FRAME 



BACK ROLL I" DIA. 



26- 
CHANQE 



-56 



FRONT ROLL I Ve DIA.- 1.125" 
FIG. 13. 

From the above sketch we figure the draft as fol- 
lows : 
\y%" front roll = 1.125 inches. Back roll = 1 inch. 

28 33i 

i.25 x n x m 

= 7.343 draft. 



11 13 
change 
gear 



192 



Then, by leaving out the change gear,. 26, we have 
a number, 26 X 7.343 = 190.918, which is the con- 
stant for draft, and 

constant 190.918 

= 26 change gear ; 



and, 



draft 7.343 



constant 190.918 
= 7.343 draft. 



change gear 26 

Example : If a 1.80 H. R. comes from interme- 
diate, and a 24 teeth change gear is used on fine 
frame, what will be the H. E. No. in front of fine 
frame ? 

constant 190.918 

= 7.95 draft. 

change gear 24 

Then if the H. R. behind is doubled its resulting 
No. is 1.80 -j- 2 = .90 H. R. 

Then .90 X 7.95 = 7.155 H. R. No. on fine frame. 

Example: What change gear should be used to 

make 4.45 H. R. from 1.80 hank on intermediate? 

1.80 -V2 = .90 H. R. resulting H. R. fed 

after doubling; 

4.45 -=- .90 = 4.94 draft ; then, 

constant 190.918 

= 38.6 or 39 draft gear. 

draft 4.94 

Example : If a 7 H. R. is to come from fine frame 
which has a 36 teeth change gear, what H. R. must 
come from the intermediate? 



193 



constant 190.918 



= 5.3 draft 



change gear 36 
then, 
H.R. delivered (7) X doubling (2) 



draft 5.3 
DRAFT OF A SPINNING FRAME 



= 2.63 H.R. 
from inter- 
mediate. 



Using the above sketch to the left for finding the 
draft between front and back rolls we have : 
dia. back roll %" = .875 dia. front roll = 1" 



= 12 draft. 



1 X 84 X 75 

.875 X30 X 20 
change 
gear 

Then if change gear, 30, is left out we have 
30 X 12 = 360 as a constant, and 



constant 360 



change gear 30 



= 12 draft ; 



or, 



constant 360 



draft 12 



= 30 change gear. 



194 



To find draft between front and middle rolls, use 
sketch to the right, and we have, 

1 X 100 X 94 
== 11.55 draft. 



.875 X 30 X 31 

change 
gear 

Leaving out change gear, 31,, we have a constant, 
'11.55 X 31 == 358.05. 
Then, 

constant 358.05 
= 11.55 draft ; 



or, 



change gear 31 
constant 358.05 



31 change gear. 



• draft 11.55 

To find the draft between the back and middle 
rolls, multiply together constant for whole draft by 
foot end change. Then multiply the constant for 
front and middle rolls by head end change. Divide 
the latter into the former. 

Example : 358.05 -f- 29 == 12.34 draft between 
front and middle rolls. 

Then, 

360 X 29 

= 1.041 draft between back and 

358.05 X 28 middle rolls. 

Both of the drafts multiplied together will give 
the total draft in machine. 

Then, 12.34 X 1.041 = 12.845 total draft. 

The above calculations and sketches are for such 
frames as the Saco-Pettee in which the "Break" 



195 



draft or draft between middle and front roll must 
be kept constant with the total draft. This "Break", 
draft does not have to be calculated for every 
change of total draft. It is only necessary to know 
that the "Break" change gear must be 1 tooth 
larger than the head end change until a 50 teeth 
gear is used on the head end. Then the "Break" or 
foot end change gear is made two teeth greater for 
all subsequent gears. 



B.R. 7 /s" = .875" DIA 



F.R I" DIA 



The above sketch, Fig. 15, is for such frames as 
the Fales and Jenks in which the middle roll is 
driven from the back roll on machine. In such ma- 
chines only one change of gears is necessary for 
changing the draft of the machine. This is found 
as follows : 
Diameter of front roll = 1". Back roll %" = .875 



10.66 draft, 



crown 
1 X 100 X 84 

.875 X 30 X 30 

change 
gear 

Leaving out the change gear, 30, we find the num- 
ber, 30 X 10.6*6 = 319.8 as a constant. Then, 



196 



and, 



constant 319.8 

= 10.66 draft ; 

draft gear 30 

constant 319.8 

=. 30 change gear. 



draft 10.66 



4n this machine the crown gear, 100, can be 
changed to produce a different draft; so it becomes 
•necessary to find a constant for each different num- 
ber of teeth in crown gear. The different sizes of 
this crown gear are usually 84, 100, 120, 138 and 147 
teeth. The constants for each of these are as fol- 
lows : 

1 X 84 X 84 
= 268.8 constant with 



.875 X 30 X (X) 84 crown gear. 

1 X 100 X 84 
= 319.8, constant with 



.875 X 30 X (X) 100 crown gear. 

1 X 120 X 84 
= 384 constant with 



.875 X 30 X (X) 120 gear. 

1 X 138 X 84 
= 441.6 constant with 



.875 X 30 X (X) 138 crown gear. 

1 X 147 X 84 
= 470.4 constant with 



.875 X 30 X (X) 147 crown gear. 

In figuring drafts on spinning frames it must be 
remembered that the resulting No. of yarn is always 



197 



heavier than the figured draft would show on ac- 
count of contraction in length of yarns due to twist- 
ing. Thus if a 6 H. R. is used behind frame, doubled 
2, and draft is 10, we would expect a yarn No. as 
follows : 

6 -f- 2 = 3 resulting No. of roving going to rolls, 
and 3 X 10= 30 yarn No. theoretical. 

If the yarn contracts in twisting after leaving the 
rolls, then the resulting No. will be less than 30. 
This contraction is usually about 3%. So if we de- 
sire to knoAv the exact No. of a yarn from the above 
draft and hank roving fed we have, taking 1 pound 
of 30 's as a basis in which there are 25200 yards, 
this will contract 25200 X -03 = 756 yards. Then 
25200 — 756 = 24444 yards in 1 pound after 
twisting. 

The No. of this would be 

24444 
= 29.1 yarn No. 



840 X 1 

Practically then we want to know what draft to 
use to produce a 30 yarn from 6 H. R. doubled 2. 
As the contraction in length is about 3% and as 
draft governs length it is only necessary to increase 
the draft 3% to produce a certain No. 

Then as 10 is only 91% of the draft required, the 
total draft is as follows : 

10 

10.32 draft to produce 30 



1.00 —.03) =.97 yarn from 6 H. R. 

doubled 2. 



198 



Using the constant 319.8 as found when using a 
100 crown gear on a Fales and Jenks frame this 
would give the following gear : 

constant 319.8 

= 30.8 or 31 change gear. 

draft 10.32 

Of course as yarns become finer, more contraction 
must be allowed on account of having more twist, 
and less must be allowed on heavier yarns. 

Use the constant 319.8 as found on a Fales and 
Jenks frame when using a 100 crown gear and solve 
the following examples : 

1. If 60 is the largest and 24 the smallest draft 
gears in stock in a mill, what is the range of drafts 1 

constant 319.8 

= 5.33 smallest draft. 

gear 60 

constant 319.8 

= 13.32 largest draft. 



gear 24 

2. Wanted a 30 yarn from 8.92 hank roving, what 
draft gear would you use ? 

8.92 -+- 2, doubling = 4.46 H. R. being fed; 
30 -=- 4.46 = 6.72 theoretical draft. 

Then allowing 3% contraction we have, 

6.72 

= 6.92 practical draft. 

(1.00— .03) ==.97 

constant 319.8 

= 46.2 or 46 change gear. 



draft 6.92 

199 



3. If a 50 teeth change gear is on frame and a 22 
yarn is required, what H. R. No. should be used? 

constant 319.8 

= 6.39 draft ; 

gear 50 

and, 

6.39 

= 6.52 actual draft allowing 

(1.00 — .02) =.98 2% contraction. 

Then, 

22 -r- 6.52 = 3.37 H. R. No. fed to frame. 
If double it would be 3.37 X 2 = 6.74 H. R. single. 

4. If a 7.00 H. R. is used doubled on frame, what 
gear must be used to make 40 's yarn? 

7 -f- 2 = 3.5 H. R. being fed ; 
r.nd, 

40 ^- 3.5 = 11.42 theoretical draft. 

11.42 

11.77 practical draft. 



(1.00 — .03) =.97 
constant 319.8 



draft 11.77 



27.1 or 27 change gear. 



200 



Twist Calculations 



Twist is put into roving to help in the handling 
of the product on following frames. Twist is put 
into yarn to cause the strand to withstand a certain 
amount of wear. It is necessary to twist cotton 
fiber to form a strong useful strand, but there is a 
limit to which fiber can be thus twisted. If this 
limit is exceeded the strength of the resulting strand 
is weakened. 

Twist in roving and yarns- is usually spoken of as 
so many ■* ' turns per inch. ' ' Thus if a yarn or roving 
is given 4 turns per inch this is the ''twist." This 
twist can be put into a cotton strand by holding one 
end and turning the other. This is done in cotton 
manufacturing by delivering the cotton fibers from 
a set of rolls (drawing rolls) and passing to a spin- 
dle on which the yarn is wound after twisting. If a 
certain roll delivers 1 inch of cotton while this spin- 
dle makes 1 turn the turns per inch would be 1. If 
the roll delivers 1 inch while the spindle makes 2 
turns, the resulting twist is 2 turns per inch, etc. 
Thus we see that the amount of twist is determined 
by comparing the amount delivered by front roll 
and the number of revolutions of the spindle in the 
same time. Again if front roll delivers 368 inches in 
One minute and spindle turns 6000 revolutions per 
minute the twist per inch would be 

6000 

= 16.3 turns per inch. 

368 

201 



From this we see that the twist is found by di- 
viding the r. p. m. of spindle by delivery of front 
roll or, 

R. P. M. of spindle 

(a) = turns per inch. 

front roll delivery in inches 

Example : The front roll turns 116 r. p. m. and is 

iy 8 inches in diameter. If the speed of spindle is 

7000 r. p. m. how many turns per inch are inserted? 

Delivery of front roll per minute = 

1.125 (iy 8 ) X 3.1416 X H6 = 409.9788 

inches per minute. 

Then from (a) 

7000 

= 17.7 turns per inch. 

409.9788 

There is a small amount of contraction in this 
twisting which affects the results thus obtained as 
well as cause the resulting yarns or roving to 
weigh heavy. This contraction usually amounts to 
about 3% and affects the resulting twist so little 
that it is usually disregarded by most spinners. 
Thus in the above if there is 3% contraction the re- 
sulting twist would be 17.7 -=- .97 = 18.2 turns per 
inch, and 18.2 — 17.7 = .5 turns per inch difference. 

As the turns per inch are found by dividing the 
speed of spindle by the delivery of front roll if we 
find the ratio of speed of spindle to the surface 
speed of front roll we will also have the turns per 
inch. Thus if the spindle makes 20 turns while the 
roll is delivering 1.25 inches, the twist would be 
20 -^ 1.25 = 16 turns per inch. Use the following 
sketch and find twist : 



202 



F.R. t'4"= 1.25'OIA. 



-56 



-30 
CHANGE 



■Z6 



^z 



FIG. 16. 



N 



As diameter of front roll is 1.25 inches it will de- 
liver 1.25 X 3.1416 X 1 = 3.927 inches for 1 turn. 
Now if we can find the number of turns the spin- 
dle makes while front roll makes 1 turn we can 
find the twist from (a). 

If roll makes 1 turn then the shaft, A, will make 



1XH5 



40 



2.875 turns. 



203 



Then shaft, B, will make 



4 

2.875 X %£ 

= 2.3 turns. 

5 



Shaft, C, will make 
2.3 X 56 



= 2.22 turns. 



58 
Spindle, D, will make 

2.22 X 50 - 

= 4.2692 turns spindle makes while 

26 front roll turns 1 time. 

Or, by substituting, we have 

us x % x 56 x n 

(b) = 4.2692 (same as above.) 

40 X n X 58 X 26 

Now if front roll delivers 3.927 inches at each rev- 
olution and if the spindle makes 4.2692 turns while 
this length is delivered the twist would be, from (a), 

speed of spindle 4.2692 

;c) = 1.08 turns 



front roll delivery 3.927 per inch. 

"We can combine both (c) and (b) under the same 
head and we have, 

115 X 24 X 56 X 50 

= 1.08 turns per inch. 

3.927 X 40 X 30 X 58 X 26 

If 3.927 = 1.25 X 3.1416, then we will have, 
204 



115 X 24 X 56 X 50 

1.08 turns per 



1.25 X 3.1416 X 40 X 30 X 58 X 26 inch (same 

as above). 

This last equation is the one usually used in cal- 
culating twist on a machine. 

If we desire to show this so as to be applicable to 
any machine we have the following diagram : 

"(F.R.gear) X (driven gears multiplied together) 

(circ.F.E,.)X (driver gears multiplied together) 
(dia. X 3.1416) twist. 

In such calculations the gear 56 on shaft B, looks 
to be a driver, but we must remember that we are 
finding the ratio of speed between the spindle, D, 
which is always constant, and the front roll. Then 
for all purposes we may neglect the fact that shaft, 
B, is the driving shaft and consider the spindle as 
the driver. Then gears 26, 58, 30 and 40 would be 
our drivers, and gears 50, 56, 24 and 115 would be 
our driven gears. 

As the gear, 30, on main shaft is the change gear 
in this case, leaving this gear out in the equation 
would produce a number which when divided by 
whatever gear is used will give the twist. This 
number is called the "Twist Constant." 

Then to find the twist constant we would have : 

115 X 24 X 56 X 50 

; : = 32.4 twist 

1.25 X 3.1416 X 40 X (X) X 58 X 26 constant. 

Then 32.4 -~ 30 = 1.08 twist ; 
or, 32.4 ~ 1.08 (twist) — 30 change gear; 
or, 30 (change gear) X 1.08 = 32.4 constant. 
Then if constant is known,, we have, 

205 



constant 

= twist; 

change gear 

or, 

constant 

= change gear. 

twist 

Or if we can find the exact twist which is pro- 
duced by a certain gear we can find the constant 
thus: 

change gear used X twist produced from 
this gear = constant. 

From the above we also deduce these main facts : 
the spindle speed remaining always the same, the 
twist is changed by changing the speed of front roll 
or delivery. Hence, more twist means a slower front 
roll speed and less production, and less twist a faster 
front roll speed and more production. We also see 
that a larger gear will run the front roll faster and, 
hence, give less twist. A smaller gear will run the 
front roll slower and give more twist. Thus we 
see that the number of teeth in gear is inversely pro- 
portioned to the twist. 

Example : If a 30 teeth gear produces 2 turns per 
inch, what gear will produce 3 turns? 

We would expect a smaller gear, so we have, not 
old gear : old twist : : new gear : twist required, but 
the following : 

old gear : twist required : : new gear : old twist., 
Then 30 : 3 : : (X) : 2 ; 

30X2 
and — - — = 20 new gear to use. 



206 



TWIST ON SLUBBER 



Back Roll 1" Draft Gear 




DRAFT AND TWIST DIAGRAM 
FIG. 17 

The above, Fig. 17, being a sketch of the gearing 
on the Saeo-Pettee frame for 12 x 6 inch bobbin the 
twist is found as follows : 

Front roll 1% 6 inches diameter, circumfer- 
ence = 1 3 / 16 X 3.1416 = 3.731". 



= 1.2495 twist per inch. 



130 X 46 X 50 X 55 

3.731 X 71 X 40 X 46 X 27 
change 
gear 

Then if we leave out the change gear, 40, we will 
have a number 40 X 1.2495 = 49.98 which is the 
twist constant ; and, 



207 



or, 



constant 49.98 '.. ...",. ; 

===== 1.2495 twist per inch; 

change gear 40 

constant 49.98 

40 change gear. 



twist 1.2495 



The usual rule for twist in roving is 1.2 X VH. R. 

No. 

In some cases where 1 inch to 1% inch staple is 

being used 1 X VH.R.No. is used on the slubber. In 
the following examples 1.2 X VlLR may be used to 
illustrate the processes : 

Examples : 1. A .62 H. R. is to be made on slub- 
ber. What twist gear must be used? 

Vl52~= .0787;- 
then 1.2 X -0787 = .944 twist per inch. 

constant 49.98 

— - — ■ — = 52.9 or 53 twist gear. 

twist .944 

2. A .75 H. R. is to be made on slubber. "What 
twist gear should be used? 

V775~= .0866. 
Then 1.2 X .0866 = 1.0392 twist per inch. 

constant 49.98 

== 48 twist gear. 

twist 1.0392 



208 



TWIST ON INTERMEDIATE 



Draft Gear 




DRAFT AND TWIST DIAGRAM 
FIG. 18 

The Fig. 18 being a sketch of gearing for the 
Saco-Pettee intermediate frame with 10 x 5 inch 
bobbin, the twist is found as follows : 

Front roll 1% 6 inches diameter. Circum- 
ference = 1% 6 X 3.1416 = 3.731". 



1.4645 twist per inch. 



130 X 39 X 42 X 44 

3.731 X 71 X 30 X 35 X 23 

change 
gear 

If the change gear,. 30, is left out we have a num- 
ber, 30 X 1.4645 = 43.935, as the twist constant. 
Then, 



constant 43.935 
twist gear 30 



1.4645 twist; 



or, 



209 



constant 43.935 

= 30 twist gear. 

twist 1.4645 



Using the above constant and 1.2 X VBL R. as 
the twist we solve the following problems : 

1. If a 1.80 EL R. is wanted on intermediate, what 
twist gear should be used? 

VT80 = 1.341. 
Then 1.341 X 1.2 = 1.6092 twish per inch; and, 

constant 43.935 

= 27.3 or 27 twist gear. 

twist 1.6092 

2. If a 2 H. R. is to be made on intermediate, what 
twist gear should be used? 

\/lT= 1.414. 
Then 1.414 X 1.2 = 1.6968 twist per inch. 

constant 43.935 

= 25.8 or 26 twist gear. 

twist 1.6968 



210 



TWIST ON FINE FRAME 



raft Gear 




DRAFT -AND TWIST DIAGRAM 
FIG. 19 

Fig. 19 being a sketch of the Saco-Pettee fly frame 
using an 8 x 4 inch bobbin the twist is found as fol- 
lows : 

Front roll 1%" diameter, circumfer- 
ence = 1% X 3.1416 = 3.534. 

130 X 39 X 53 X 44 

= 2.0694 twist per inch. 

3.534 X 71 X 30 X 33 X 23 
change 
gear 

Leaving out the change gear, 30, we find a num- 
ber, 30 X 2.0694 = 62.082, twist constant. 
Then, 

constant 62.082 

= 30 twist gear ; 



and, 



twist 2.0694 



constant 62.082 



twist gear 30 



= 2.0694 twist per inch. 



211 



Using 1.2 X V H. R. as the twist per inch and the 
above constant we solve the following problems: 

A 5.46 H. R. is wanted on fine frame. What twist 
gear must be used? 

V~5\46 = 2.337. 
Then 1.2 X 2.337 = 2.8044 twist per inch, and 
constant, 

62.082 

= 22 twist gear. 



twist 2.8044 

If a 7.00 H. R. is wanted on fine frame, what twist 

gear must be used? 

V7.00 = 2.645. 
Then 1.2 X 2.645 = 3.174 twist per inch, and 

constant 62.082 

= 19.55 or 20 twist gear. 

twist 3.174 

TWIST ON A SPINNING FRAME 

In ring spinning the actual twist is put into the 
yarn by the speed of the traveler as compared with 
the delivery of the front roll. The "lag" of the 
traveler behind the spindle to wind the yarn on 
the bobbin will give the speed of the traveler. Thus 
if a spindle is running 8000 r. p. m. and has a bobbin 
% of an inch in diameter while front roll is deliver- 
ing 480 inches per minute the traveler must "lag" 
behind the spindle a sufficient number of turns in 
order to wind on this 480 inches of yarn,, which 
would be as follows : Circumference of bobbin = 
% X 3.1416 = 2.3562 inches. Then the bobbin must 
have 1 revolution more than the traveler to wind 
on this 2.3562 inches, and to wind on 480 inches it 



212 



will have, 480 -^ 2.3562 = 203.7 more turns per 
minute than the traveler. This means that the 
traveler will have 203.7 turns less than the spindle 
which is 8000 — 203.7 = 7796.3 turns per minute. 

This means that if the twist per inch is figured by 
comparing the speed of the spindle to the delivery 
of front roll, the actual twist is less than the twist 
thus figured. Now as there is a certain amount of 
contraction due to twisting this one thing will cause 
more twist per inch than will be figured and will 
almost offset the loss in twist due to the difference 
in speeds of travelers and spindles. 

Therefore, for all practical purposes the speed of 
the spindle compared to the delivery of front roll 
will give the twist per inch. 



20- 

TWIST 



50- 



-60 








CYLINDER 


7" 


) 




DIA. / 








WHIRL 3 A'' DIA. 
FIG. ZO. 



213 



In figuring twists on spinning frames a number is 
used called "Ratio of Cylinder to Whirl." This 
ratio simply means the number of turns the whirl or 
spindle turns while the cylinder is making one turn. 
This number can be found on any machine by put- 
ting a mark on the whirl and by turning the cylin- 
der slowly for, say, 5 turns and counting the num- 
ber of turns the spindle makes. The total turns 
made by the spindle divided by the number of turns 
the cylinder makes gives the ratio. 

For example, on foot end of frame mark with 
chalk a point on both frames and cylinder coincide 
ing. Then mark also a point on the whirl and some 
close part coinciding. Then turn the cylinder until 
both marks again come together at both places. 

Example : Cylinder makes 13 turns while whirl 
makes 107 before all marks again come together. 
Then 107 -=- 13 = 8.23 ratio. 

Using the gearing as shown in Fig. 20 and a ratio 
of cylinder to whirl of 8.24 the twist is found as 
follows : 

1 inch front roll. 
Circumference = 1 X 3.1416 = 3.1416 inches. 

ratio 
60 X H2 X 8.24 

= 17.6255 twist per inch; 

3.1416 X 50 X 20 

cylin- twist 

der gear 

gear 

or leaving out the twist gear, 20, we have, 

60 X 112 X 8.24 

= 352.51 which is the same as 

3.1416 X 50 

214 



twist, 17.6255 X 20, (twist gear) =352.5 twist 
constant; then, 

constant 352.51 

= 20 twist gear; 

twist 17.6255 

or, 

constant 352.51 



= 17.6255 twist per inch. 

twist gear 20 

The twist per inch for any number of warp yarn 
is found by multiplying the square root of the num- 
ber by 4.75, or 

4.75 X VNo. = turns per inch for warp. 
The twist per inch for any number of filling is 
found by multiplying the square root of the num- 
ber by 3.5 ; or,, 

3.5 X VNo. = turns per inch for filling. 

It is well to remember that the more twist a ma- 
chine has the slower the speed of front roll and 
hence less production. Also the less twist, the faster 
the front. roll and more production. 

Examples : Use constant as found above and 
solve. 

1. If a 40 filling is to be made, what twist gear 
should be used? 

Vlo = 6.324 
Then 6.324 X 3.5 = 22.13 turns per inch. 



constant 352.51 
twist 22.13 



15.9 or 16 twist gear. 



2. A 30 warp is to be made. "What twist gear 
should be used? 



215 



V30 = 5.477. 
Then 5.477 X 4.75 = 26.01 turns per inch. 

constant 352.51 

= 13.5 or 14 twist gear. 

twist 26.01 



216 



Production Calculations 



Production on any machine is based on three 
things ; namely, speed of delivery roll, circumfer- 
ence of roll and weight of each yard or certain num- 
ber of yards. Thus, if a roll is 9 inches in diame- 
ter and is turning 6 r. p. m. it will deliver, 

9 X 3.1416 = 28.2744 inches for 1 turn, and 
28.2744 X 6 = 169.4464 inches in 1 minute, or 

169.4464 

= 4.706 yards in 1 minute. 

36 

As there are 60 minutes in 1 hour, this will deliver, 

4.706 X 60 = 282.360 yards in 1 hour. 
If this material weighs 12 ounces to the yard, we 
will have, 
282.36 X 12 = 3388.32 ounces delivered in 1 hour. 
Then, 3388.32 ~ 16 (ounces to 1 lb.) = 211.77 
pounds delivered in 1 hour. 
If the total time run is to be considered for 60 
hours, then, 

211.77 X 60 = 12706.20 pounds made in 60 hours. 
Such calculations may be put under one head by 
placing all numbers multiplied above a line and all 
used in dividing below a line ; thus, 

9 X 3.1416 X 6 X 60 X 12 X 60 

= 12706.2 pounds. 

36 X16 

In all such calculations it is customary to pick 
out all items which are constant and produce a 

217 



number known as a "production constant." In the 
above example the only variable quantities would 
be the speed of roll and the weight per yard. Then 
leaving these two out we would have, 

9 X 3.1416 X 60 X 60 

= 176.419 production con- 

36 X 16 stant for 60 hours. 

Then to find the production we would multiply 

the speed of delivery roll by wxight in ounces per 

yard of product and multiply this by 176.419. Thus, 

6 X 12 = 72 ; and 

176.419 X 72 = 12706.2 pounds production. 

An allowance of time in stopping caused by un- 
avoidable reasons is usually made. Then the re- 
duction as found above is allowed in forming the 
constants for a certain time. 

Such figured productions on machines are called 
"Theoretical Productions," and means the max- 
imum possible production or 100%. In figuring the 
"per cent of production" for a machine or mill this 
theoretical production is taken as a basis or max- 
imum and the actual production in pounds weighed 
is compared to this. For example, if the above 
12706.2 pounds is the theoretical production and we 
find that 10862 pounds are produced, the % pro- 
duction would be, 

10862 X 100 

■ = 85.4% production. 

12706.2 

PRODUCTION OF A PICKER 

The speed of front roll or delivery roll being one 
of the factors in finding production it is necessary 



218 



to determine this first. TMs can be done in two 
ways. First we can use a speed indicator for count- 
ing this speed direct from the roll. Every over- 
seer should have one of these for finding his speeds ; 
as slippage of belts and other things will produce a 
variation from the figured speeds. Second, the 
speed of delivery roll may be figured from some 
constant driving point. 

As sufficient calculations have been made to show 
how speeds are figured only the resulting speed of 
the delivery roll will be considered. 

Allowing 10% for loss by stoppage the production 
on a picker having 9 inch lap rolls running 6 r. p. m 
and making 14 ounce laps would be for 60 hours as 
follows : 

9 X 3.1416 X 6 X 60 X 60 X 14 X -90 

= 13307.5 

36 X 16 pounds. 

If we leave out the variable item, 6, speed of rolls 
and 14, weight of lap, we find the production con- 
stant as follows : 

9 X 3.1416 X 60 X 60 X .90 

■ = 158.42 production 

36 X 16 constant. 

Then to find production multiply together speed 
of lap roll and weight per yard and multiply this by 
the constant, 158.42. 

Example : If a 12 ounce lap is being made, what 
would be the production in 60 hours if roll is run- 
ning 6 r. p. m. ? 

12 X 6 = 72 ; 
158.42 X 72 = 11406.24 pounds. 



219 



Example : If in above example only 10462 pounds 
are produced, what is the % production? 

10462 X 100 

= 91.7% production. 

11406.24 

PRODUCTION OF A CARD 



"3 


LICKER-IN 


< 

< 
— a 


20— 


CYLINDER 


< 
5" 

00 

190- 


- DOFFER 27" 





FIG. 21. 

220 



Fig. 21 being a sketch for the driving of the 27 
inch doffer on Saco-Pettee card, the speed is found 
as follows: 

R.P.M. of cylinder 165. 

change 
165 X 18 X 4 X 20 

= 8.81 R.P.M. of doffer. 

7 X 18 X 214 

Leaving out the change gear, 20, we have a con- 
stant as follows : 

165 X 18 X 4 

= .4405 speed constant. 

7 X 18 X 214 

Then speed constant, .4405 X change gear, 20 = 
8.81 speed. Also speed, 8.81 -i- speed constant, 
.4405 = change gear 20. 

Although the production should be figured from 
coiler calendar rolls it is usually figured direct from 
the doffer. Using a 27 inch doffer and 8.81 as rev- 
olutions per minute on a 56 grain sliver the pro- 
duction in 60 hours, allowing for 10% stoppage, 
would be as follows : 

27 X 3.1416 X 8.81 X 60 X 60 X 56 X -90 

= = 527.3 

36 X 7000 lbs. in 

60 hrs. 

Leaving out the variable numbers, 56 and 8.81, 
the weight of sliver and speed of doffer respectively, 
we have, 

27 X 3.1416 X 60 X 60 X -90 

= 1.069 production 

36 X 7000 constant. 

221 



Then, speed of doffer X grain sliver X constant = 
pounds production; or, speed of doffer = 

pounds production 



grain sliver X constant 

Example: If 624 pounds production is required 
on a card making a 54 grain sliver, what production 
gear must be used? 

First, speed of doffer = 

624 

= 10.15 R.P.M. of doffer. 

54 X 1.069 

Then as speed constant as seen above for doffer is 
.4405 we have, 
speed, 10.15 -=- constant, .4405 = 23 production gear. 

Example : If a 54 grain sliver is being made with 
a doffer speed of 9.25 R.P.M., what is the production 
in 60 hours? 

constant = 1.069. 

Then constant, 1.069 X 9.25 X 54 = 533.96 pounds 
production. 

Example : If 498 pounds of the sliver in example 
2 is produced, what is the % production ? 

498 X 100 

= 93.2% production. 

533.96 

PRODUCTION OF A DRAWING FRAME 

In figuring production on a drawing frame one 
delivery is taken as a basis and about 10% loss al- 
lowed for stopping. 

Example : What is the production in 60 hours of 



222 



a drawing frame if the front roll is 1% inches in di- 
ameter, making 400 revolutions per minute on a 50 
grain sliver ? 

If a metallic roll is used its diameter, if a 16 pitch, 
would be figured as 1.66". If a 24 or 32 pitch its 
diameter would be 1.50". 

1.66 X 3.1416 X 400 X 60 X 60 X 50 X-90 

- = 1340.22 

36 X 7000 lbs. in 

60 hrs. for 
one delivery. 

If we leave out the speed of front roll and weight 
of sliver we have a constant thus, 

1.66 X 3.1416 X 60 X 60 X-90 

= .06701 produc- 

36 X 7000 tion constant. 

Then,, production constant X speed of front roll X 
weight in grains of sliver = pounds production in 60 
hours. 

PRODUCTION OF ROVING FRAMES 

As the production on slubbers, intermediates and 
fine frames is calculated the same way one example 
will be given to illustrate all. The allowance for 
time lost depends upon the size of H. R. being made 
as heavier roving will make more sets in a day ne- 
cessitating more doffing. 

As the speed of front roll varies for every differ- 
ent size of roving it is not advisable to. use the speed 
of this roll in finding the theoretical production. 
The best method is to base the calculation on the 
speed of spindles, which is constant, and the twist 



223 



per inch, which is always known for any roving 
made. 

Example : A fine frame is making 7 H. R. with a 
spindle speed of 1200 E.P.M. What is the produc^ 
tion per spindle in 60 hours, allowing 10% stoppage? 
Speed of spindle divided by twist per inch gives 
inches delivered in a minute. 

^J~= 2.645, and 2.645 X 1.2 = 3.174 turns per inch. 
Then 1200 -=- 3.174 = 378.07 inches delivered per 
minute. 

And 378.07 X 60 X 60 = 1361052 inches deliv- 
ered in 60 hours. 

1361052 -j- 36 = 37807 yards in 60 hours. 
And, 

37807 

= 6.4 pounds per spindle in 60 hours. 

7 X 840 

In order to find the amount for a frame of 160 
spindles we have, 

6.4 X 160 = 1024 pounds. 

We can put all of these calculations under one 
head and allowing for 10% stoppage we have, 

1200 X 60 X 60 X -90 

= 5.76 pounds per spindle 

3.174 X 36 X 840 X 7 in 60 hours. 

The following will give a production constant : 
1200 (spindle speed) X60X6OX-90 



= 128.57 pro- 

36 X 840 duction con- 

stant for 60 hrs. 
Then, 

constant 

= lbs. per spindle for 60 hours. 

twist X counts 

224 



The actual production of a frame is shown in 
hanks (840 yards) delivered by the front roll. This 
will give the production for 1 spindle and must be 
multiplied by whatever number of spindles is con- 
sidered. 

Example : If clock registers 41.5 hanks of 7 H. R. 
for 60 hours, what is the pounds production per 
spindle ? 

As 1 hank = 840 yards, then 41.5 hanks would 
be 41.5 X 840 = yards, 

as the weight is found by dividing yards by 
No. X 840 we have, 

41.5 X 840 41.5 

= = 5.92 pounds per spindle. 

7 X 840 7 

Rule to find production from hank clock: Divide 
numbers of hanks run by H. R. No. 

Example : If a 5.46 H. R. is being made with 1200 
spindle speed, what would be the production in 60 
hours for 1 spindle? 

V5\46 = 2.337 
Then 2.337 X 1.2 = 2.804 twist. 
Using the constant as found above we have, 

128.57 

= 8.39 pounds per spindle in 60 hrs. 



2.804 X 5.46 theoretical production. 

If we have 160 spindles to a frame and 10 frames 
this would be 160 X 10 X 8.39 = 13424 pounds 
made on 10 frames in 60 hours. 

Example : If hank clocks on 10 frames register 
a total of 441 hanks, on 5.46 H.R. as above, what in 
the actual pounds production on the 10 frames? 



225 



441 

= 80.77 pounds actual production on 10 

5.46 spindles,, as this registers for 1 

spindle on each frame, 

or, 80.77 -=- 10 = 8.077 pounds actual production 

on 1 spindle for 60 hours. Then 

8.077 X 10 X 160 = 12923.2 pounds 

production actual. 

If we compare this with the theoretical production 

of 13424 we have, 

12923.2 X 100 

= 96.2% production. 

13424 

PRODUCTION OF A SPINNING FRAME 

The production of a spinning frame can be found 
as on the fly frame; namely, by using the spindle 
speed and twist of yarn as a basis. 

Example : What is the production of 1 spindle in 
60 hours on 30 's warp yarn if the spindle speed is 
9500 R.P.M., allowing a loss of 10% for stoppage? 
4.75 X V~30~ = 26.02 twist. 

Then, 

9500 X 60 X 60 X -90 

1.464 pounds per spin- 



26.02 X 36 X 30 X 840 die in 60 hours. 

Leaving out the variable numbers, 30, No. of yarn, 
and twist, 26.02, we find a constant for 60 hours at 
a spindle speed of 9500. For different spindle speeds 
other constants must be worked out. 

Thus, 



226 



9500 X 60 X 60 X-90 

— ■ = 1142.79 production con- 

36 X 840 stant for 1 spindle 

for 60 hours. 

Then, 

constant 

: = pounds per spindle in 60 

twist X No. yarn hours. 

Thus, 

1142.79 

= 1.464 pounds per spindle in 60 

26.02 X 30 hours on 30 warp. 

Example : With a spindle speed of 9500 R.P.M., 
what would be the production in 60 hours of 30000 
spindles on 28 's warp yarn? 

4.75 X V~28~= 25.13 twist. 
Then, 

constant 1142.79 

= 1.623 pounds production for 

25.13 X 28 1 spindle in 60 hours. 

Then, 

1.623 X 30000 = 48690 pounds production, 

theoretical production. 

As the % production is based on the number of 
spindles run rather than on the total spindles in a 
mill it is necessary to keep a record of spindles 
stopped every day and from this find the total spin- 
dles run in a week. 

Example : If a spinning room has 12 sides of 128 
spindles stopped on Monday, 10 sides on Tuesday, 
8 sides on Wednesday, 4 sides on Thursday, 10 sides 



227 



on Friday and 10 sides for a half day on Saturday, 
what is the total number of spindles run in a week? 
We have 5% days in which the total number of 
spindles stopped would be : 

Monday 12 X 128 = 1536 spindles 

Tuesday 10 X 128 = 1280 spindles 

Wednesday 8 X 128 = 1024 spindles 

Thursday 4 X 128 = 512 spindles 

Friday 10 X 128 = 1280 spindles 

Y 2 Saturday 10 X 128 = 1280 spindles 

5% days 6912 spindles 

Then, 

6912 

= 1256.7 or 1257 average spindles 

5% stopped for week. 

If the mill has 30000 spindles, the total spindles 
run would be 30000 — 1257 = 28743. 

Using 1.623 as the pounds production per spindle 
we have, 

28743 X 1-623 = 46649.8 theoretical produc- 
tion in 60 hours or 1 week. 

If the actual production is 45826 pounds, then, 

45826 X 100 

= 98.2% production. 



46649.8 
PRODUCTION OF LOOMS 

The basis for figuring production on a loom is 
that of picks per inch put into cloth and speed or 
picks per minute. 

Example : If loom is making 160 picks per min- 



228 



lite and producing cloth with 64 picks per inch, how- 
many yards will it make in 60 hours? If cloth 
weighs 4.75 yards per pound, how many pounds will 
be made in 60 hours? 

160 -v- 64 = 2.50 inches made per minute ; 
2.5 X 60 = 150 inches made in 1 hour; 
150 X 60 = 9000 inches made in 60 hours, 
and, 9000 -j- 36 = 250 yards in 60 hours. 

This may be put in one process by placing all 
numbers multiplied above a line and those used for 
dividing below a line. Then, 

160 X 60 X 60 

== 250 yards made. 

64X36 

This is 100% production. Then, 

250 -+- 4.75 = 52.63 pounds. 

If the speed of the looms remain the same, the 
only variable quantity in the above is the picks per 
inch, 64. Then if we leave this out we have 

160 X 60 X 60 

— = 16000 production constant. 



36 

constant 



= yards production in 60 



picks per inch hours. 

Example : If looms run 160 picks per minute with 
72 picks per inch,, what is the yards production in 
60 hours? 

16000 

= 222.22 yards in 60 hours. 

72 

229 



The speed of a loom is best found by counting 
the picks made in a minute. The size of pulleys to 
produce a certain number of picks per minute is 
found as follows : 

Example : Line shaft runs 200 R.P.M., pulley on 
loom 12 inches in diameter ; what size pulley must 
be used on line shaft to produce 160 picks per 
minute ? 

200 X (X) = 12 X 160 

12 X 160 

X = = 9.6 or 9^o inch pulley. 

200 

As this figures 9.6 and we use a 9% inch pulley, it 
is best to find at what speed the loom will run using 
this 9% i nc h pulley. As line shaft is making 200 
R.P.M. we have, 

9y 2 x 200 = 12 X (X) 

9y 2 X 200 

and (X) = = 158.3 picks per minute. 

12 

To find the % production of looms: As produc- 
tion is figured on actual number of looms run per 
week rather than on total looms, it is necessary to 
find first the average number stopped. This may be 
done as follows : 

Example : A mill has 865 looms on 72 x 68 4.60 
yards goods 36" wide. If looms run 160 picks per 
minute, find % production from following : 



230 



Number looms stopped per week 

Monday 3G 

Tuesday 72 

Wednesday 98 

Thursday 42 

Friday 98 

Y 2 Saturday 98 

5% days 444 

and 444 -f- 5% = 80 average looms stopped. 

Then 865 — 80 == 785 total looms run. 
160 X 60 X 60 

. = 16000 production constant 

36 for 60 hours. 

Then, 

16000 

= 235.29 yards for 1 loom for 60 hours ; 

68 

and, 

235.29 

= 51.15 pounds per loom for 60 hours. 

4.60 

Now 785 X 235.29 = 184702.65 yards theoretical 
production for 785 looms. This is 100%. 

And 51.15 X 785 = 40152.75 pounds theoretical 
production for 785 looms. This is 100%. 

If the actual production in yards is 181216 yards, 
the % is as f ollows : 

181216 X 100 

= 98.9% production. 

184702.65 

The actual production in yards being 184702 this 
should be 184702 ~ 4.60 = 40152.6 pounds of cloth. 



231 



If it weighs more than this the mill is losing money 
provided the cloth is sold on a yard basis. Suppose 
that the weight of the above production is 41321.9 
pounds. 

Then we are putting in 41321.9 — 40152.6=1169.3 
pounds more cotton than is necessary,, which could 
have been used in making 4.60 X 1169.3 = 5378.78 
yards more of cloth. If this cloth were worth 12 
cents per yard we would be losing, 

5378.78 X -12 = $645.45. 

To find the % of seconds: If in making the above 
181216 yards (actual production) Ave make 2500 
pounds of seconds, this would be 11500 yards of 
cloth. Then, 

11500 X 100 

: = 6.3% seconds. 

181216 

If these seconds are sold for 10 cents per yard 
they would bring 11500 X .10 = $1150.00. If they 
had all been turned into good cloth they would 
have brought 11500 X -12 = $1380.00. This is a loss 
of $1380.00 — $1150.00= $230.00 outside of the ad- 
ditional expense of preparing and marketing. 

The above calculations are made without any al- 
lowance for stopping. This would be correct on 
draper looms, but for non-automatic looms .10% 
should be allowed for stoppage. 



232 



Cloth Calculations 



MAKE-UP OF CLOTH 

The make-up of a piece of cloth embodies the 
weight per lineal yard, the width, the construction 
(ends per inch and picks per inch) and the weave. 
As the weight for a certain construction is determ- 
ined from the size of warp and filling used, the de- 
termination of the yarn numbers are of importance. 

WEIGHT OF CLOTH 

The weight of a cloth is always (except in very 
heavy construction such as duck) determined by the 
number of yards contained in 1 pound. Thus a 
"4 yards goods" has 4 yards in 1 pound; a "5.30 
yards goods" has 5.30 yards in 1 pound and so on. 
To find the weight of a piece of cloth, a certain 
length, say, 60 yards, can be taken and weighed and 
the yards per pound found. For instance, if 60 
yards of cloth weighs 13% pounds the weight would 
be, 60 -=- 13% = 4.44 yards per pound. If a mill 
sells its cloth by the yard based on so many yards 
per pound it is of great importance that the weight 
does not run heavy. If it does the mill is losing 
money both in cotton and excess freight paid. 

Thus, if a cloth is to weigh 4.25 yards per pound 
and it is made up to have only 4.15 yards per pound 
the amount of loss in cotton by weight would be as 
follows : 

Taking 60 yards as a standard length, the weight 



233 



of this length should be 60 -=- 4.25 = 14.117 pounds. 

If, however, it weighs only 4.15 yards per pound, 
the weight of 60 yards would be, 60 -j- 4.15 = 14.455 
pounds. 

Then 14.455 — 14.117 = .338 pounds loss for each 
60 yards. If a mill makes 600,000 yards of this, the 
total loss in pounds would be as follows : 

60 yards : .338 pounds :: 600,000 yards : X pounds; 

.338 X 600000 

then X = = 3380 pounds loss. 

60 

WIDTH OF CLOTH 

The width of a piece of cloth from the loom de- 
pends on how wide the warp is set in the reed and 
the % of shrinkage caused by the interlacing of the 
-Jthreads. Thus if a warp is set 38% inches in reed 
and has 8% shrinkage we will have, 38.5 X -08 = 
3.08 inches shrinkage which will make cloth, 
38.5 — 3.08 = 35.42 inches wide. The width in the 
reed will be determined from the reed number used 
and the number of ends in the warp together with 
the number of ends put in each dent. Thus if a 
number 32 reed (32 dents per inch) is used and 
warp has 2460 ends drawn 2 per dent, the width is 
found as follows : 

2460 -^ 2 = 1230 total dents to use. 
Then 1230 -=- 32 (reed No.) = 38.43 width in reed. 

If we have the width of cloth given and know the 
% shrinkage we can find the width in reed. 

Example : If cloth is to be 34 inches wide and 
has 7% shrinkage, what must be width in reed? 



234 



If it has 7% shrinkage then 34 inches is 100 — 7 = 
93% of the width in reed or 1 inch in reed will 
shrink to .93 inch in cloth. Then we have the pro- 
portion : 

1 inch in reed : .93 inch in cloth : : X inches 
in reed : 34 inches cloth ; 

or 1 : .93 : : X : 34. 
.93 X X = 1 X 34; 

1X34 

then X = = 36.55 inches in reed. 

.93 

From this we have the rule to find width in reed. 
Divide the width of cloth by 100 — % shrinkage 
(expressed in decimals). 

If, in the above example the cloth is to have 60 
ends per inch drawn 2 per dent, the reed number is 
found as follows : 

34 (width of cloth) X 60 (ends per 
inch) = 2040 ends in warp. 
2040 -r- 2 (ends per dent) = 1020 dents in reed. 
Then as width in reed is to be 36.55 inches we 
have, 

1020 (dents) -f- 36.55 (width in reed) = 27.9 
dents per inch, or reed number. 
This is practically a 28 reed. So if 28 is used the 
actual width in reed is found by dividing this 28 
into the total dents, 1020. 

Thus, 1020 -^ 28 = 36.42 width in reed. 
It is a hard thing to determine the exact amount 
of shrinkage that a cloth is going to have in weav- 
ing as there are so many things which affect it. 
Some of these are as follows : 



235 



Tension on Warp 

The tighter a warp is run on the loom, the nar- 
rower the cloth will be, and the slacker it is run the 
wider it will be. It is a known fact that a change 
of from Y 8 to y 4 of an inch and sometimes more, can 
be made on a loom by thus changing the tension of 
the warp. This is taken advantage of by designers 
in selecting the % of shrinkage a cloth will have. 
They judge from experience about the % to. allow 
for a certain cloth and any discrepancy can be made 
up on the loom as given above. 

Height of Sand Roll 

The position of the sand roll has some effect on 
the amount of shrinkage. A high sand roll, such as 
used on a Draper loom, will cause the cloth to have 
a smaller shrinkage than a low sand roll, such as is 
used on a Crompton and Knowles or Whitin loom. 
The difference is usually about 2%. That is, if a 
cloth would have a shrinkage of 9% on a Whitin 
loom, it will have only about l°/o on a Draper loom. 

Sizes of Yarns Used 

A cloth using heavier yarns will have more % 
shrinkage than one having finer yarns provided the 
ends and picks per inch in both are the same. 

Ends and Picks Per Inch 

The number of ends and picks per inch has an 
effect on the % of shrinkage. A cloth having a 
smaller number of ends per inch will have less % 
shrinkage than one having a larger number of ends, 
provided the size of yarn is the- same. 



236 



The Weave Used 

A plain weave will have more °/o of shrinkage than 
any other, provided the construction and yarn sizes 
are the same. A three harness drill will have less 
than a plain weave. A four harness weave less than 
a three harness, and so on. 

CONSTRUCTION OF CLOTH 

When we speak of the "Construction" of a cloth 
we usually mean its make-up as regards warp ends 
and picks. Thus, the construction of certain cloth 
may be 60 warp ends per inch and 54 picks per inch. 
This is usually shown thus, 60 X 54. 

If the cloth were made having the same number 
of ends drawn to each dent it is only necessary to 
count the number of ends contained in 1 inch of 
cloth in finding its construction. If, however, the 
warp were draAvn having more ends in some dents 
than in others it is necessary to find what is known 
as the "Over all Construction." This is done by 
counting the number of ends in one repeat of the 
pattern and measuring several repeats and comput- 
ing from these. Thus if a pattern has 98 ends in 1 
repeat and 6 repeats measures 5.25 inches the "over 
all ' ' is found as follows : 

98 X 6 = 588 total ends on 6 repeats = 

5.25 inches. Then 588 -^ 5.25 = 112 

over all construction. 

As different numbers of yarns have different di- 
ameters it is necessary to find these diameters so 
as not to use more ends in one inch than can lie side 
by side in 1 inch. Thus, if we have a yarn which is 

237 



Y 8 of an inch in diameter it is clear that 8 of such 
ends would lie side by side in one inch and no more. 
If we have a yarn which is % 6 of an inch in diame- 
ter, 16 of such ends will lie side by side in 1 inch, 
and so on. 

There are two things which determine the great- 
est numbers of ends per inch in any cloth. These 
are : first, the diameter of the yarn, and, second, 
the weave used. 

Taking the plain weave as an example, the fol- 
lowing sketch shows a cross section of cloth made 
with this weave. ■ 




FIG.22. 

In Fig. 22 we show the ends of one set of threads, 
a, b, c and d. Interlacing with these according to 
the plain weave is the thread, X. In this case we 
see that the thread, X, passes between each of the 
ends, a, b, c and d and hence must separate them 
from each other a distance equal to the thickness, 
or diameter, of the thread, X. If the diameter of 
the ends, a, b, c and d, are y 8 of an inch each, then 
8 of such ends would lie side by side in 1 inch. But 
as the thread, X, separates these ends one from the 
other a distance equal to its diameter it is clear that 
8 of the ends, a, b, c and d, cannot be placed into one 



238 



inch of such cloth. If the diameter of X is also % 
of an inch the spaces between the ends, a, b, c and d, 
must be % of an inch to allow this thread to pass. 
Then the spaces marked 1, 2, 3 and 4 will include 
the diameter of ends a, b, c and d -j- the diameter 
of X. This space will then be % of an inch and only 
four of the ends, a, b, c and d, can be put in one 
inch of such cloth. 

Diameter of Yarn 

The diameter of any number of yarn can be found 
if we know its density or the weight of such yarn 
in a cubic inch or cubic foot. About 28% cubic 
inches of cotton fiber compressed together as they 
are in warp twisting will weigh 1 pound. As No. 1 
yarn has 840 yards = 840 X 36 inches in 1 pound, 
the volume contained in this length of No. 1 yarn 
is 28.25 cubic inches. Then as volume = (diame- 
ter) 2 X .7854 X length we have, 

28.25 (cubic inches) = (diameter) 2 X -7854 
X 840 X 36 ; solving the equation we have, 

28.25 

= (diameter) 2 

.7854 X 840 X 36 

or, as .7854 X 36 = practically 28.25 we cancel 

and have % 40 = (diameter) 2 . 
And by finding square root of both sides we have, 

= diameter. 



Or, 

= diameter of No. 1 = % 9 of an inch. 



V840 

239 



As 840 is the number of yards in 1 pound of No. 1, 
then to find the diameter of any yarn we have. 

1 

= diameter of any No. 



V number yds. in 1 lb. 

Or as 840 X No. yarn = number yards in 1 pound. 
If we find the square root of 840 and divide into 
1 we have, 



V 840 = 28.9 and = .034. 

28.9 



Then we have, 

.034 



= diameter in inches. 



V No. 
Warp and Filling- Numbers 

The construction, weight and width being given 
to make a piece of cloth it becomes necessary to find 
the sizes of warp and filling to produce this weight. 
If we bear in mind that a yarn No. takes into con- 
sideration length and weight we see that we must 
know the length of yarn required to make a certain 
piece of cloth as well as the weight in order to find 
the Nos. of yarn. 

Average Number 

If nothing is given but ends and picks per inch, 
weight and width of cloth, it is usually the practice 
to find average No. of yarn from the total weight 
of cloth and the length of yarn both warp and 
filling in this weight. From this average No., the 



240 



proper warp and filling Nos. can be found. Suppose 
we have to make a cloth. 36% inches wide, 72 ends 
and 64 picks per inch and to weigh 4.80 yards per 
pound. If we consider this in 840 yard lenghts the 
calculation will be more accurate and easier. 

Thus, if cloth is to weigh 4.80 yards per pound 
840 yards will weigh 840 -f- 4.80 = 175 pounds. 
(a) 72 X 36.5 = 2628 ends in warp. 

Selvage 

If 36.5 inches is the width of the cloth it is the 
width over selvage and all. Then 2628 will include 
selvage ends. If we use 12 ends of doubled selvage 
on each side the total selvage would be 24 ends. 
This would have 2628 — 24 = 2604 ends in body 
of warp. But as these 24 ends are doubled we ac- 
tually use 48 single ends to make them. Then 
2604 + 48 = 2652 total ends in a warp, or the sel- 
vage allowance is found by adding to the ends as 
found under (a) the number of ends in both sel- 
vages. Thus 2628 + 24 = 2652 ends in warp. 

Then the total yards of yarn in the warp for 1 
yard would be, 

2652 ~ .93 (1.00 — .07) = 2851 yards. 

This allows for 7% take-up. 

The yards of filling would be as follows : 64 picks 
X 36.5 cloth width = inches of filling in 1 inch of 
cloth. 

64 picks X 36.5 X ffl 

■ = 2516 yards of filling in 

ffl X -93 1 yard of cloth. 

Then 2516 + 2851 = 5367 yards total yarn in 1 
yard cloth; and 840 X 5367 = total yards in 840 
yards of cloth. Then as 

241 



number yards 

— = No., 

840 X lbs- 

£40 X 5367 

we have = 30.6 average No. 

X175 



If the number of yards of filling and warp were 
the same, then if a 30.6 warp were used a 30.6 fill- 
ing would have to be used. But this will not be 
true where there is a difference in the yardage of 
warp and filling. If this were true then if we de- 
sire to have 10 numbers difference between our warp 
and filling we would take a 25.6 warp and a 35.6 
filling, because 

25.6 + 35.6 
= 30.6. 

2 

What would be the weight if we use this 25.6 
warp and 35.6 filling? 

£40 X 2516 

= 70.674 pounds filling in 840 yards 

£40 X 35.6 cloth ; 

and 

X 2851 111.367 lbs. warp in 840 yds. cloth 



£40 X 25.6 182.041 lbs. in 840 yds. cloth 

This would be a difference of 182.041 — 175 = 
7.041 pounds loss on every 840 yards,, which figures 
about 4%. 

When, therefore, we desire to make a difference 
between the warp and filling numbers, we must de- 
crease the average number to find the warp number 



242 



only as much as the difference between one-half 
this number and the ratio between warp and filling 
yardage. Thus in the above example if we desire 
to make a difference of 10 numbers between the 
warp and filling, one-half of 10 = 5 and ratio of 
warp to. filling is 

2851 

= 1.113. 

2516 

Then 5 — 1.113 = 3.87 and 30.6 — 3.871 = 26.713 
which will be our warp number and 36.7 will be the 
filling number. 

Then to find the weight produced with these num- 
bers we have, 

X2516 

= 68.88 pounds filling in 840 yards. 



X 36.7 
And, 

X 2851 106.78 lbs. warp in 840 yds. 



X 26.7 175.66 lbs. in 840 yds. cloth. 

If the number of warp be given to find the fill- 
ing, then taking the above example if a 28 's warp is 
to be used we have, 

X2851 

= 101.81 pounds of warp in 840 yds. 



X28 

In the above calculations we have made no allow- 
ance for size on the warp. In this case then if the 
warp has 8% of size, the total weight of warp in 840 
yards would be, 



243 



101.81 

= 110.66 pounds weight of warp 



.92 (1.00 — .08) in 840 yards. 

840 yards of cloth = 175 pounds, then 175 — 
110.66 = 64.34 pounds of filling in 840 yards. 

$40 X 2516 

Then = 39.1 filling No. 

^X 64.34 
If a 40 filling is desired to be used in the above ex- 
ample, then to find warp numbers we have, 

$40 X 2516 

= 62.9 pounds of filling in 840 yards; 

$40 X40 

and, 175 — 62.9 = 112.1 pounds of sized warp 

in 840 yards. Then 112.1 X .08 (size) = 8.96 

pounds of size in 112 pounds of warp, and 

112.1 — 8.96 = 103.14 pounds of warp 

before sizing. Then 

$40 X2851 

27.6 warp No. 



$40X103.14 

To find the weight of cloth from a small sample : 
If only a small sample of cloth is to be had its weight 
can be found by cutting a piece a certain length 
along the warp and filling edges, thus forming a 
square or rectangular piece. This sample is ac- 
curately weighed in grains and the weight obtained 
as follows : 

Example : Cloth is to be 36% inches wide and a 
small sample 6x6 inches weighs 54 grains ; how 
many yards would there be in 1 pound? It is not 
absolutely necessary to take a sample 6x6 inches, 
but it is best to do so if it can be obtained. 



244 



Sample 6 x 6 = 36 square inches. 

1 yard of cloth 36% inches wide = 36 X 36.5 

square inches. Then 

36 X 36.5 

= 36.5 of the small samples would be 

36 contained in 1 yard. 

If sample weighs 54 grains then 1 yard of cloth 
would weigh 36.5 X 54 = 1971 grains. 
Then, as 7000 grains = 1 pound,, we have, 
7000 -f- 1971 = 3.55 yards per pound. 

Example : Make all necessary calculation for 
making a cloth 36 inches wide 68 x 60 and to weigh 
4.75 yards per pound with a 30 's warp having 8% 
size. "Weave plain. (Allow 7% shrinkage.) 
36 X 68 = 2448 ends in warp. 

With 12 selvage ends the allowance extra is 24. 

Then, 2448 + 24 = 2472 total ends in warp for 
weight. 

Then, 2472 -j- .93 (1.00 — .07) = 2658 yards of 
warp in 1 yard. 

36, width of cloth, -f- .93 = 38.7 inches width in 
reed. Reed 2 per dent. 

2448 -;- 2 = 1224 total dents to use in reed. 

1224 -r- 38.7 = 31.6 reed number (dents per inch). 

If two cotton harness are to be used each harness 
must have 

2448 -=- 2 = 1224 eyes spread on 38.7 inches. 

38.7 (width in reed) X 60 (picks) = 2322 yards 

filling in yard of cloth. Allowing for a length 
of 840 yards of cloth we have, 



245 



(yds. warp in 1 yd.) 
= 88.6 lbs. of 



f>fty X 30 yarn No. warp in 

840 yds. 

58.6 

= 95.6 pounds of sized warp in 



.92 (1.00 — .08) 840 yards. 

If cloth is to weigh 4.75 yards per pound 840 yards 
will weigh 840 -^ 4.75 = 176.84 pounds. 

Then, 176.84 — 95.6 = 81.24 pounds of filling in 
840 yards, and 

$40 X 2322 yards filling 

: = 28.5 filling No. 

jB£0 X 81.24 



246 



Pra&ice and Cost Finding 

By practice is meant the method of finding the 
total cost of materials when rated at so much a 
pound, yard, etc. Cost finding is the reverse of prac- 
tice in that it is the method of determining the cost 
per pound, yards, etc., after including all materials 
going into the manufacture of that article. 

Examples : 1. A mill buys 400 bales of cotton 
weighing on an average of 510 pounds each. The 
price paid is 25 cents per pound. What is the 
total cost? 

400 X 510 = 204000 pounds of cotton ; 
then 204000 X $0.25 = $51000 cost. 

If the bagging and tires weigh 24 pounds per 
oale, what is the amount paid ? Note, these are paid 
for at the same rate as the cotton. 

24 X 400 = 9600 pounds of bagging and tires ; 
and 9600 X $0.25 = $2400 paid for bagging 
and tires which is included in $51000 above. 

If, when sold, they bring $2400, the actual cost 
of the cotton would be $51000 — $2400 = $48600.00. 

But if they bring only 12 cents per pound,, we 
would get only 9600 X -12 = $1152.00' for them. 

Then they would cost the mill $2400 — $1152 = 
$1248.00. 

Then the actual cost of the resulting weight of 
cotton, 204000 — 9600 = 194400 pounds, would be 
$51000.00 + $1248.00 = $52248.00. 



247 



To find the cost per pound in cents, multiply the 
total cost by 100 and then divide by the pounds. 
Thus, to find the actual cost per pound of the 
194400 pounds of cotton which cost a total of 
$52248.00, as we have seen, we have, 

52248.00 X 100 

= 26.87 cents per pound. 

194400 

2. A mill runs in a week, 500,000 pounds of raw 
cotton and makes 12% waste. How much cloth 
should be made? 

500000 X .12 = 60000 pounds waste. 
Then it should make 500000 — 60000 = 440000 
pounds of cloth. 

If the raw cotton cost 26 cents per pound and the 
waste is sold at an average of 12 cents, what is the 
cost per pound of cotton going into cloth? 

60000 pounds of waste which was paid for at 26 
cents per pound would cost 60000 X -26 = $15600. 
If sold at 12 cents we receive back 60000 X -12 = 
$7200. Then we must add the difference, $15600 — 
$7200 = $8400, to the cost of the cotton put into 
cloth. Thus, 

440000 X -26 = $114400.00 ; 

and 114400 + 8400 = $122800.00 cost of 

440000 pounds of cotton. Then, 

122800 X 100 

= 27.9 cents per pound. 

440000 

3. What would be the cost of 216 looms at $185 
each? 

216 X $185 = $39960. 



248 



4. At $40 per 1000 (usually indicated by letter, 
M), what would be the cost of 50000 wire heddles? 

If price is by the thousand, cut off three figures 
to the right of the number indicating the amount 
and then multiply. Thus, 

50.000 X 40 = $2000. 

5. At $15 per M (1000) what would be the cost 
of 11650 travelers? 

11.650 X $15 = $174.75. 
At $12 per hundred what would be the cost of 
2652 picker sticks? 

When the price is by the hundred cut off two fig- 
ures by decimal point to. the right of number show- 
ing the amount. Then, 

26.52 X 12 = $318.24. 

6. If cloth is sold at 12 cents per yard and it 
Weighs 4.48 yards per pound,, what is the selling 
price per pound ? 

4.48 X -12 = 53.76 cents per pound. 

7. A bale of cloth weighing 520' pounds contains 
cloth weighing 5.25 yards per pound and is worth 
15 cents per yard. What is the bale worth? 

520 X 5.25 = 2730 yards in the bale; 
and 2730 X -12 = $327.60. 

8. The price paid for 1 horse power per year is 
around $20. If a mill has 840 looms requiring 1 
H. P. for every 3 looms, what should be the weekly 
cost for power? 

840 -i- 3 = 280 horse power required. 
This would be 280 X 20 = $5600 per year, or 
$5600 

= $107.69 per week for 840- 

52 weeks in a year looms. 



249 



If these looms produce 42000 pounds per week, 
what is the power cost per pound? 

107.69 X 100 

= .256 cents. 

42000 

The above, .256 cents, is a fraction of a cent or 
about % of a cent. In showing cost per pound this 
is the best method to use. It is sometimes shown 
with the dollar mark. Thus, in the above, if it is 
to be shown expressed in dollars we would not have 
used the number 100, to multiply the number 
$107.69. When we use 100, as there are 100 cents in 
a dollar, we reduce the dollars to cents and the re- 
sult is in cents or fractions thereof. 

9. A card room produces 46824 pounds of rov 
ing per week. The weekly payroll is $513.85. What 
is the cost per pound of labor? 

513.85 X 100 

= 1.09 cents per pound. 

46824 

10. A man runs two roving frames and makes 162 
hanks in a week at 8 cents per hank. What is his 
pay? 

162 X .08 = $12.96. 

11. A weaver running 16 looms makes 62 cuts per 
week at 21 cents per cut. What is his pay? 

62 X -21 = $13.02. 

12. A cloth is to be made which will give 3% cuts 
per week on a loom, allowing an average of 16 looms 
per weaver. How much should be paid per cut so 
that weavers can make $16.75 per week? 

16 looms X 3% cuts = 56 -cuts per week. 



250 



16.75 X 100 

Then = 29.9 cents or 30 cents per 

56 cut to be paid. 

13. A cut of cloth has 8.46 pounds of warp valued 
at 42 cents per pound, 6.57 pounds of filling at 39 
cents per pound. What is the cost of yarn in 1 cut? 

8.46 X -42 =. $3.5532 cost of warp. 

6.57 X -39 = $2.5623 cost of filling. 

$6.1155 total cost of warp and filling. 

14. If a cloth weighs 4.30 yards per pound and 
has 65% warp in it valued at 50 cents per pound, 
what is the cost per yard if filling is worth 42 cents 
and cloth has 60 yards per cut? 

60 -f- 4.30 = 13.953 pounds, weight of 1 cut. 
13.953 X -65 = 9.069 pounds of warp in 1 cut. 
9.069 X -50 = $4.5345 cost of warp in 1 cut. 
13.953 — 9.069 = 4.884 pounds of filling in 1 cut. 
4.884 X -42 = $1.8512 cost of filling in 1 cut. 
4.5345 + 1.8512 == $6.3857 cost of 1 cut of 60 yards. 

6.3857 X 100 

Then = 10.64 cents per yard. 

60 

15. A speeder tender runs two frames of 160 
spindles each and produces 98.8 hanks of No. 7 H. R. 
at 8 cents a hank. What is the cost per pound for 
labor ? 

1 spindle on each frame will have made 
98.8 

= 49.4 hanks. 

2 

Then 49.4 -f- 7 H. R. = 7.057 pounds per spindle. 
Then 160 X 2 = 320 total spindles. 



251 



And 320 X 7.057 = 2258.24 pounds production on 
320 spindles. 

98.8 X -08 ■— $7,904 paid for producing 2258.24 
pounds. Then 

7.904 X 100 

= .35 cents per pound. 

2258.24 

This can also be found as follows : 

98.8 X -08 X 100 X 2 X 7 



= .35 cents per pound. 

320 X 98.8 

Then to find the cost per pound for any price paid 
per hank regardless of the amount of production 
we have, 

price paid in cents X 100 X H.R. No. 

= cost per 

number spindles per frame pound. 




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